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What a function approaches, not what it equals
Say you want one number for how steep a curve is — not on average over some stretch, but exactly at one point. The slope between two points is easy: rise over run. Squeeze the second point down until it sits on the first, and the run becomes zero — rise over zero, undefined, nothing to divide. Most of us would stop there and call the question unanswerable.
Try it anyway on $f(x) = (x^2-1)/(x-1)$ at x=1. Direct substitution gives $0/0$ — no answer. But simplify first: for any x other than 1, $(x^2-1)/(x-1) = x+1$. As x gets closer to 1 from either side, $x+1$ gets closer to 2. The function is not even defined at x=1, yet the value it is heading toward is perfectly clear.
That heading-toward value is a limit. A limit asks what value a function APPROACHES as its input approaches some point — not what the function equals AT that point, which may be undefined, different, or the same value. The derivative, this chapter’s real subject, is built entirely on that distinction:
$f'(x) = \lim_{h \to 0} (f(x+h) - f(x))/h$
evaluates a ratio undefined exactly at h=0. It is the limit, not a direct substitution, that gives it a value there.
Read every claim in this chapter for what the function approaches, never for what it equals.
One picture, one instance of the whole chapter’s opening idea. The curve keeps climbing toward 1 no matter which side you approach from, while the function’s actual value sits somewhere else entirely, unconnected to that climb. Every derivative in this chapter leans on that same gap between heading-toward and landing-on.
Sometimes “the function approaches something” has two different answers depending on which side you arrive from. Picture a function that jumps at x=a: coming in from the left it climbs toward 3, but coming in from the right it climbs toward 5. Which one is the limit at a?
Neither, on its own. A limit at x=a exists only if two one-sided limits agree. The LEFT-HAND LIMIT ($\lim_{x \to a^-} f(x)$, x approaching a from below) must exist. The RIGHT-HAND LIMIT ($\lim_{x \to a^+} f(x)$, x approaching a from above) must exist too, and equal the first. When both hold, that common value is the limit.
If the two one-sided limits differ, or either fails to exist, the two-sided limit does not exist at that point. This holds regardless of what $f(a)$ itself is — even if f is not defined there at all.
Check both sides before writing down an answer: a single substitution never tells you whether the two sides agree.
One test decides it: do the left-hand and right-hand limits agree?
Two functions, two outcomes, one test applied identically both times: follow each arrow to where it is heading, then compare only those two destinations against each other. Nothing else about either function enters the verdict.
Worked example
Compute $f'(1)$ for $f(x) = x^2$, from the definition
$f'(1) = \lim_{h \to 0} [(1+h)^2 - 1]/h$ Form the difference quotient directly from the definition.
$(1+h)^2 - 1 = 2h + h^2$ Expand the numerator.
$[2h + h^2]/h = 2+h$ Divide by h — legal for any $h \neq 0$, exactly the regime the limit lives in.
$\lim_{h \to 0} (2+h) = 2$ Let h go to zero.
slope $= 2$ at $(1,1)$ That 2 is the SLOPE of the tangent line to $y=x^2$ at (1,1) — the limiting slope of the secant through (1,1) and $(1+h, (1+h)^2)$ as h shrinks to zero. Every derivative-at-a-point computation in this chapter repeats this same shape: form the difference quotient, simplify away the h, then let h go to zero. The number a derivative returns is always a slope first, a rate of change wearing whatever units the problem gives it.
Three real secant slopes — 4, then 3, then 2.5 — are not approximations pulled from a formula; each one is the actual output of the difference-quotient arithmetic at one concrete h. Run the same arithmetic for a smaller h and the number keeps landing nearer 2. This is exactly why the algebra later in this chapter is trusted to skip straight to a limit instead of computing case by case.
Building limits from simpler limits — the algebra of limits
▲CONCEPT
Working every limit from the raw definition — forming a difference quotient, simplifying, letting h shrink to zero — would make even a simple polynomial limit painfully slow. Four laws let you skip that, PROVIDED the pieces you are combining already have limits of their own.
Given $\lim_{x\to a} f(x) = l$ and $\lim_{x\to a} g(x) = m$, both already existing, four laws follow.
SUM: $\lim_{x\to a} [f(x)+g(x)] = l+m$.
DIFFERENCE: $\lim_{x\to a} [f(x)-g(x)] = l-m$.
PRODUCT: $\lim_{x\to a} [f(x) \cdot g(x)] = l \cdot m$, which covers the scalar case $\lim_{x\to a} [c \cdot f(x)] = c \cdot l$ for constant c.
Each law needs BOTH limits to already exist — the laws build a new limit out of two known ones; they do not create a limit where none exists. Miss that proviso and a law gets applied where it does not hold.
Four laws, one condition on every single one of them: the pieces must already have limits before you can combine them.
The green curve was drawn from its own formula, not by adding the other two on the page. So the fact that it passes through 3 exactly where the other two pass through 1 and 2 is a result, not a construction. That is the whole content of the first three laws: a combination’s limit can be assembled from the pieces’ limits without going back to the definition. The fourth law is the one with an extra condition, and every 0/0 later in this chapter is that condition failing.
Worked example
Evaluate $\lim_{x\to 2} [3x^2 + 5x - 1]$ by the algebra of limits
$\lim_{x\to 2} [3x^2 + 5x - 1] = \lim_{x\to 2} 3x^2 + \lim_{x\to 2} 5x - \lim_{x\to 2} 1$ Split it into three pieces by the sum and difference laws — legal only because each piece is already known to have a limit.
$\lim_{x\to 2} 3x^2 = 3 (\lim_{x\to 2} x)^2 = 3(2^2) = 12$ Apply the scalar and product laws.
$\lim_{x\to 2} 5x = 5(2) = 10$ Apply the scalar law again; the constant term stays 1.
$12 + 10 - 1 = 21$ Add the pieces back — every step used one of the four algebra-of-limits laws, never a new rule, never a fresh appeal to the definition.
Two routes to the same number, drawn to the same scale so the agreement is visible rather than asserted. The stack is built in the order the laws release the pieces. 3 times the limit of x squared, then 5 times the limit of x, then a constant taken off. It finishes exactly level with the bar that was never taken apart. Nothing is gained by splitting this particular expression; direct substitution reaches 21 in one step. What the picture buys is confidence in the method for the cases later in the chapter where substitution has no answer at all.
Every algebra-of-limits law from the last section has one job: build a new limit from ones already known to exist. Polynomials and rational functions are exactly where that job pays off immediately.
Take any polynomial f. It is built entirely from sums, products, and scalar multiples of x — nothing else. Apply the algebra-of-limits laws to each of those operations in turn, all the way down. The result is always $\lim_{x\to a} f(x) = f(a)$ — direct substitution works, every time, for every polynomial.
A rational function $f(x)/g(x)$ extends the same result one step further, through the quotient law: $\lim_{x\to a} f(x)/g(x) = f(a)/g(a)$, PROVIDED $g(a) \neq 0$. That proviso is not fine print — it is the quotient law’s own condition from the last section, carried forward untouched. The moment $g(a)=0$, the quotient law no longer applies, and direct substitution is not merely inconvenient; it is undefined.
Direct substitution is not a shortcut you reach for — it is what the algebra of limits already guarantees, right up until a denominator vanishes.
Same kind of function, same point-by-point test, two different outcomes — and the test is checkable before you substitute anything: does the denominator survive at that point? That one question sorts every rational limit into one of these two panels.
▲CONCEPT
A denominator vanishing does not always mean the limit fails — it can mean the algebra hasn’t been simplified far enough yet.
Take a rational limit $\lim_{x\to a} f(x)/g(x)$ where BOTH $f(a)=0$ and $g(a)=0$ — the 0/0 indeterminate form. Direct substitution is undefined here, but the limit can still exist. Since f and g both vanish at x=a, the factor theorem guarantees that $(x-a)$ divides both of them exactly. Factor it out of numerator and denominator, cancel the shared $(x-a)$, and substitute into what remains.
This exact technique proves a genuinely general result. For any positive integer n,
$\lim_{x\to a} (x^n - a^n)/(x-a) = n a^{n-1} \cdot$
The factorisation $x^n - a^n = (x-a)(x^{n-1} + x^{n-2}a + \ldots + a^{n-1})$ cancels the $(x-a)$ directly, leaving n terms that each equal $a^{n-1}$ once $x=a$ is substituted — n copies of $a^{n-1}$, added together.
The same 0/0 shape sits behind average speed itself. Distance over time becomes exactly 0/0 the instant the time interval shrinks to zero — which is the whole reason a derivative needs a limit rather than plain division.
A 0/0 answer is not a dead end — it is a signal to factor first, substitute second.
Both denominators vanish at x = 1, and that alone decides nothing. On the left the numerator vanishes with it — 0 over 0. So the shared factor cancels and what is left is an ordinary line missing one point, which the curve still heads toward from both sides. On the right the numerator is 1 and cancels nothing, so the quotient grows without bound and there is no height to head toward. Reaching for factoring is not optimism: 0 over 0 is the signal that a factor is there to cancel, and 1 over 0 is the signal that none is.
Worked example
Evaluate $\lim_{x\to 1} (x^3-1)/(x^2-1)$
at $x=1$: $0/0$ Direct substitution gives $0/0$ on both counts — factor first.
$x^3-1 = (x-1)(x^2+x+1)$; $x^2-1 = (x-1)(x+1)$ Factor numerator and denominator.
$\lim_{x\to 1} (x^2+x+1)/(x+1)$ Cancel the common $(x-1)$.
$(1+1+1)/(1+1) = 3/2$ Now direct substitution works, because the denominator no longer vanishes at x=1.
Worked example
Evaluate the same limit by the standard form
$((x^3-1)/(x-1)) / ((x^2-1)/(x-1))$ Divide top and bottom by the same $x-1$.
$\lim_{x\to 1} (x^3-1)/(x-1) = 3 \times 1^2 = 3$ The standard form $\lim_{x\to a} (x^n-a^n)/(x-a) = n a^{n-1}$, with $n=3$, $a=1$.
$\lim_{x\to 1} (x^2-1)/(x-1) = 2 \times 1^1 = 2$ The same form again, $n=2$.
$3/2$ Same answer as factoring — two routes, one limit.
The worked example divides top and bottom by the same (x-1) and then takes two limits, 3 and 2. This is what that permission looks like. The orange curve is the indigo one divided by the violet one at every x, so its height is their quotient everywhere, and the height it heads for is their quotient too. All three circles are open, because none of the three has a value at x = 1 at all.
Polynomials and rationals fold into limits already known by direct substitution or factoring. Trigonometric functions near zero do neither — they need one new geometric fact, stated once and reused everywhere.
With x measured in RADIANS (not degrees — the geometric proof ahead depends on it), three standard limits hold.
$\lim_{x\to 0} \sin(x)/x = 1$.
$\lim_{x\to 0} \tan(x)/x = 1$ as well, and it follows directly from the first: $\tan(x)/x = [\sin(x)/x] \cdot [1/\cos(x)]$, and $\cos(0)=1$.
$\lim_{x\to 0} (1-\cos(x))/x = 0$ as well: $(1-\cos(x))/x = 2 \sin^2(x/2)/x = [\sin(x/2)/(x/2)] \cdot \sin(x/2)$, which tends to $1 \cdot 0 = 0$.
None of these three come from direct substitution — each one starts as $0/0$ at x=0 — and none come from the algebra-of-limits laws alone. They need a geometric argument, worked through below.
Three limits, one geometric proof behind all of them — stated here so the proof itself has something to aim at.
Two of these curves close on the same height from opposite sides and the third closes on a different height altogether, which is the one distinction worth carrying out of this section. The reason sits in the numerators. sin x and tan x both shrink at the same rate as x itself, so their ratios settle at 1. 1 - cos x shrinks far faster and its ratio settles at 0. Three formulas would have to be memorised separately; three curves are one picture.
▲CONCEPT
Why does $\sin(x)/x$ actually approach 1? Not by algebra — by comparing three areas on the unit circle.
For $0 < x < \pi/2$, picture the unit circle with angle x. Comparing the areas of two triangles and the circular sector trapped between them gives $\sin(x) < x < \tan(x)$. Divide through by $\sin(x)$, which is positive in this range, and take reciprocals — reversing both inequalities: $\cos(x) < \sin(x)/x < 1$.
As $x \to 0$, $\cos(x) \to 1$. $\sin(x)/x$ is sandwiched between $\cos(x)$ and 1, both closing in on 1 from either side, so $\sin(x)/x$ has nowhere else to go — it must approach 1 too. This is the SANDWICH theorem: trap an unknown quantity between two known ones heading to the same place, and it is forced there as well.
The same argument, run for $x \to 0^-$, gives the identical result — $\sin(x)/x$ is an even function of x, so the negative side mirrors the positive side exactly.
Three quantities, one squeeze — the limit has no room to be anything but 1.
Every piece of this drawing is load-bearing. The three shaded areas are not illustrating the inequality, they ARE the inequality; measuring them is the entire proof that sin(theta)/theta has nowhere to go but 1.
Worked example
Evaluate $\lim_{x\to 0} \sin(3x)/x$
$\lim_{x\to 0} \sin(3x)/x$ The standard result is $\sin(x)/x \to 1$, but here the argument of sine is $3x$, not x — match them first.
$3 \cdot [\sin(3x)/(3x)]$ Rewrite by multiplying and dividing by 3, so the same quantity $3x$ sits in both the sine and the denominator.
as $x \to 0$, $3x \to 0$ too The bracket approaches 1 by the standard trig limit.
$3 \cdot 1 = 3$ Whenever the argument inside sine does not match the denominator, force them to match first — multiply and divide by whatever it takes.
Neither curve has a value at x = 0. Both are 0/0 there - so both open circles are genuinely empty, and the only question either picture answers is where the curve is heading. What differs between them is the angle fed to the sine. A reader who applies the standard limit straight to sin(3x)/x is answering the lower picture while looking at the upper one. The rewrite in the table is what moves the question from one to the other.
Building derivatives from simpler derivatives — the algebra of derivatives
▲CONCEPT
The limit definition $f'(x) = \lim_{h\to 0} [f(x+h)-f(x)]/h$ computes any derivative directly. Re-deriving it from scratch for every function every time is exactly as slow as it sounds. Four rules, each proved once from that same definition, let a derivative be built from simpler derivatives instead.
Given $u=f(x)$ and $v=g(x)$, both differentiable.
SUM/DIFFERENCE: $(u \pm v)' = u' \pm v'$.
PRODUCT: $(u \cdot v)' = u' \cdot v + u \cdot v'$.
SCALAR MULTIPLE: $(c \cdot u)' = c \cdot u'$ for constant c.
The product and quotient rules are the two to watch. They are NOT $(u \cdot v)'=u' \cdot v'$ and NOT $(u/v)'=u'/v'$ — multiplying or dividing the two derivatives directly is the single most common algebra slip this toolkit invites. Each rule is proved by applying the algebra-of-limits laws to the difference quotient of the combined function — the same machine that built the limit toolkit builds this one too.
Four rules to reach for, and exactly one trap: never multiply or divide the derivatives directly for a product or a quotient.
The dashed line is what happens if the product rule is remembered as f' times g'. It passes through the right point, so nothing about it looks wrong until it is drawn: at 0.21 it is almost flat where the curve is climbing steeply. The solid line, from u’v + uv’, lies along the curve for as far as it is drawn. Sum and difference really do split term by term; product and quotient never do, and this is one measured instance of the size of that difference.
Both lines pass through the same point with the same steepness, so neither is obviously wrong on its own. The error shows only against the curve, and only because the curve is falling. That is the practical warning: a sign slip in this rule’s numerator produces an answer that survives every check except comparing it with the function it came from. Writing v squared underneath is not what makes the rule work; writing u’v first is.
Worked example
Differentiate $f(x) = x^2 \sin(x)$
$u = x^2$, $v = \sin(x)$ It is a product of two functions, so the product rule applies directly rather than expanding anything by hand.
$u' = 2x$ The same pattern the general $(x^n-a^n)/(x-a) \to n a^{n-1}$ result already produces, named properly as the power rule next section.
$v' = \cos(x)$ The derivative of sine, proved from first principles in the next section — taken here as a preview, the way a rule is often used once before it is formally derived.
$f'(x) = u'v + u v' = 2x \sin(x) + x^2 \cos(x)$ The product rule gives two terms, one from differentiating each factor while holding the other fixed — never $u'v'$ alone.
Read the gold curve first. It is positive while the cosine is, crosses zero exactly at x = pi/2, and is large and negative after that. This is what drags the solid curve down through zero near x = 2.5, where the original function x² sin x stops rising. The violet curve never turns negative anywhere in this window. So the answer’s two terms are not two halves of one thing: one of them sets how big the slope is, the other decides when the slope changes sign.
Worked example
Differentiate $f(x) = x/(x+1)$
$u = x$, $v = x+1$ Both are linear, so both derivatives come straight from the definition.
$[(x+h)-x]/h = h/h = 1$, so $u' = 1$ The difference quotient for x itself — even this obvious derivative comes from the same definition as everything else in this chapter.
$v' = 1$ The same reasoning gives $v'=1$ for $x+1$ — the added constant cancels in the difference quotient.
The four algebra steps beside this graph can only be checked against themselves. Here the answer is checked against something else: at x = 1 the violet curve reads 0.25, and the green line drawn tangent to f at that same x also has slope 0.25. Read the pair across the whole window and the relationship keeps holding. f climbs steepest near the origin, exactly where the violet curve is highest, and flattens as the violet curve falls away.
Derivatives of polynomial and trigonometric functions
$d/d x (x^n) = n x^{n-1}$, for any positive integer n — this is the general pattern the last two worked examples already leaned on, now stated and derived properly.
The derivative definition gives $\lim_{h\to 0} [(x+h)^n-x^n]/h$. Substitute $y = x+h$, so $y-a=h$ with $a=x$, and $y \to a$ exactly as $h \to 0$. The limit becomes $\lim_{y\to a} (y^n-a^n)/(y-a)$ — the exact general result the same factoring technique proves: $n a^{n-1}$. With $a=x$, that is $n x^{n-1}$.
One substitution turns the derivative definition into a limit already known — the power rule is not a new fact, it is the old one renamed.
The 2 is a count of strips, not a copy of the exponent — a square has two directions to grow in, so two strips survive and the corner does not. Grow a cube the same way and three slabs survive, which is 3x². The square proves n = 2 and the cube n = 3; the algebra in the text proves every n.
▲CONCEPT
$d/d x \sin(x) = \cos(x)$ — the product-rule example that already used this fact gets its proof here, from the definition directly.
From the limit definition,
$d/d x \sin(x) = \lim_{h\to 0} [\sin(x+h)-\sin(x)]/h \cdot$
The sum-to-product identity $\sin(A)-\sin(B) = 2 \cos((A+B)/2) \sin((A-B)/2)$, with $A=x+h$ and $B=x$, turns the numerator into $2 \cos(x+h/2) \sin(h/2)$. Dividing by h and regrouping: $\cos(x+h/2) \cdot [\sin(h/2)/(h/2)]$.
As $h \to 0$: $\cos(x+h/2) \to \cos(x)$, and $\sin(h/2)/(h/2) \to 1$ by the standard trig limit proved earlier this chapter. What remains is exactly $\cos(x)$.
The identical method, using $\cos(A)-\cos(B) = -2 \sin((A+B)/2) \sin((A-B)/2)$ in place of the sine identity, gives $d/d x \cos(x) = -\sin(x)$.
AC current and sound waves are both modelled as sinusoids — computing how either changes moment-to-moment is exactly this limit, run on a real signal instead of an abstract x.
The same three-step shape from every derivative in this chapter — form the difference quotient, simplify, let h shrink to zero — proves this one too.
The derivation on this page ends in a symbol, and a symbol cannot be checked by eye. These three places can. Where the sine curve is climbing hardest, at x = 0, the cosine curve is at its peak. Where the sine is dead flat on its crest, the cosine is crossing zero. Where the sine is falling hardest, the cosine has bottomed out. Three chances for the answer to have been the wrong function, and it was the right one each time.
Worked example
Differentiate $f(x) = x^3 + \cos(x)$
$f(x) = x^3 + \cos(x)$ It is a sum of two terms, so the sum rule applies — differentiate each term separately, then add.
$d/d x(x^3) = 3x^2$ Apply the power rule.
$d/d x(\cos(x)) = -\sin(x)$ Apply the first-principles result just proved.
$f'(x) = 3x^2 - \sin(x)$ Add the two terms — two rules from two different sections, used together on one function; that combination, not any single formula, is the actual skill.
The gold tangent is the one to watch. It tilts downward, and it is the only one of the three that does, because the derivative of cos x is minus sin x and sin 1 is positive. Adding a term that is itself shrinking makes the sum’s tangent less steep than the cubic’s, not more - which is the whole difference between 3 and 2.159 on the page. The middle column of the table is the actual answer to the worked example; the drawing is here to say that the answer’s two pieces behave the way the tangents do.
Every result in this chapter, from the very first worked example to the last, reduces to one operation: evaluating
$\lim_{h\to 0} [f(x+h)-f(x)]/h \cdot$
Sometimes that limit is worked out once, in general, and handed over as a reusable RULE — the power rule, the algebra of derivatives, sin(x) turning into cos(x). Sometimes it is applied directly to one specific function, with no rule to reach for. Knowing which situation you are in — reach for a rule, or fall back to first principles — is the chapter’s real skill. No single formula in it matters as much as that judgment.
Economics gives this idea its own working name. The derivative of a cost or revenue function is called MARGINAL cost or marginal revenue — the field’s native vocabulary for exactly the same rate-of-change idea this chapter builds from scratch.
One operation, many disguises — everything in this chapter is that same limit, computed once and reused, or computed fresh.
Limits and derivatives run as two strands that meet once: the standard trig limits feed straight into the derivative of sine, the only link between an idea on one strand and a result on the other.Every idea here descends from one root — that a limit is about approach, not arrival — and the deepest sits seven levels below it. That depth is a reading order: the power rule stands five levels down, so five other ideas have to be in place before it can mean anything.
TRAP: if $\lim_{x\to a} f(x)$ exists, it feels like f must be defined at a, and its value there must equal the limit.
REALITY: a limit describes only the APPROACH — values of f NEAR a, taken as x gets arbitrarily close, and the definition never once looks at x=a itself. $f(a)$ can be undefined. It can be defined to a completely different value. Either way, the limit at a is entirely unaffected — this chapter’s very first example, $f(x)=(x^2-1)/(x-1)$, is undefined exactly at the point its limit describes.
This does not mean $f(a)$ is irrelevant everywhere — it means the limit and the function’s value are two separate questions, and this chapter has only ever answered the first one.
Check what f approaches. Never assume it tells you what f equals there.
The trap does not just fail once. It fails in two different ways side by side: a function with no value at all at the point, and a function with a value that is flatly wrong. Both keep the exact same limit regardless, which is the stronger proof that f(a) was never part of the calculation to begin with.
✕MISCONCEPTION
TRAP: dy/dx looks like a fraction — a top part and a bottom part — so it feels safe to split it apart and recombine the two pieces like any ordinary fraction.
REALITY: dy/dx is ONE symbol for a single limit, $f'(x) = \lim_{h\to 0} [f(x+h)-f(x)]/h$. Later algebra sometimes treats it like a fraction — the chain rule mnemonic is the most common place this shows up — but that is a convenient resemblance, not the definition. Nothing in $\lim_{h\to 0} [f(x+h)-f(x)]/h$ splits into two independently meaningful parts called dy and dx.
The notation looks like a fraction. The definition underneath it isn’t one.
The two halves look identical at a glance — same dy, same dx — but only one of them is ever actually defined. You can point to a limit for the whole symbol, never to a separate limit for its top or its bottom alone.
JEE Evaluate $\lim_{x\to 4} (x-1)/(x-4)$, or state that it does not exist.
does not exist
$1/4$
$3$
$0$
JEE Evaluate $\lim_{x\to -2} (x^3+1)/(x+5)$.
$3$
$-7/3$
$-8/3$
$1$
JEE Evaluate $\lim_{x\to 2} (x^4-16)/(x-2)$.
$0$
$64$
$32$
$24$
JEE Evaluate $\lim_{x\to -1} (x^3+1)/(x+1)$.
$0$
$-3$
$1$
$3$
JEE Evaluate $\lim_{x\to 3} (x^5-243)/(x-3)$.
$405$
$0$
$1215$
$324$
JEE Evaluate $\lim_{x\to 2} (x^2-x-2)/(x-2)$.
$0$
$3$
$4$
$1$
JEE Evaluate $\lim_{x\to -3} (x^2+2x-3)/(x+3)$.
$0$
$1$
$-4$
$-2$
JEE Evaluate $\lim_{x\to 2} (x^3-8)/(x^2-4)$.
$0$
$1$
$2$
$3$
JEE $f(x) = (x^2-1)/(x-1)$ for $x \neq 1$, and $f(1)=100$. Find $\lim_{x\to 1} f(x)$.
$100$
does not exist
$2$
$0$
JEE $g(x) = 2x+1$ for every $x \neq 3$, and $g$ is left undefined at $x=3$. Find $\lim_{x\to 3} g(x)$.
does not exist
$6$
$3$
$7$
JEE $h(x) = (x^3-8)/(x-2)$ for $x \neq 2$, and $h(2)=0$. Find $\lim_{x\to 2} h(x)$.
$12$
$0$
does not exist
$4$
Answers
$6$
$1/3$
DNE — LHL $=-1$, RHL $=1$.
$12$
$8$
$80$
$5$
DNE — LHL $=3$, RHL $=5$.
$10$
$6$
$12$
$8$
does not exist — LHL $=-1$, RHL $=1$, and they disagree
does not exist — LHL $=2$, RHL $=1$
does not exist — LHL $=1$, RHL $=2$
$12$
$5$
$8$
$-1$
$-4$
$11$
$2$
does not exist
$-7/3$
$32$
$3$
$405$
$3$
$-4$
$3$
$2$
$7$
$12$
Four leaves now, not three. A denominator that vanishes without the numerator vanishing too is not a 0/0 form. There is no shared factor to cancel. So the limit is unbounded, and the answer is DNE: exactly Exercise 12.1’s own item 23. Every limit in the exercise still ends at one of these four leaves, so reaching for a technique that is not on this chart means the question has been misread.
Every wrong option in both items is a number the question already printed, and putting them on the scale shows where each one really is. On the right, option B and the value h(2) = 0 you were handed are the same point, twelve below where the curve goes. Read the height the curve arrives at, not the height left lying around near it.
Four of Exercise 12.1’s questions are this one shape with different numbers in it. The curves show why the answer is never the fraction’s own value at x = a - there is none - and why it is not 1 either. The higher the power, the steeper the numerator vanishes, and the open circle climbs with it. The table is the arithmetic that lets a reader do the remaining three in a line each, without factorising anything.
Exercise 12.2 — Standard trigonometric limits
practice Evaluate $\lim_{x\to 0} \sin(5x)/x$.
practice Evaluate $\lim_{x\to 0} \tan(x)/x$.
practice Which of these equals $\lim_{x\to 0} (1-\cos(x))/x$?
The window is narrow on purpose. sin(7x) turns over at 7x = pi/2. This is x of about 0.22, so a picture drawn even a little wider would show this quotient falling away and would say the opposite of what is true. A limit at 0 is a claim about a small neighbourhood and nothing else. The reason these questions are answered by a rule rather than by substituting a small number is that no single small number is small enough to be safe.
The chapter’s caution box says this in one line and never draws it — in degrees the same ratio approaches pi/180, not 1. The two readings are not close, which is why the wrong option in several of these items is the whole answer scaled by pi/180 rather than a small slip.
The indigo curve passes through the origin from below, so its two one-sided limits arrive at 0 from opposite signs and still agree. This is what makes 0 a genuine two-sided answer rather than a coincidence. The violet curve is even and never reaches the axis at all. Three of the exercise’s four questions are the first picture and one is the second; the table’s middle column is how to tell which one a question is, before any working.
Miscellaneous — practice set
practice Differentiate $f(x) = x^7 - 3x^4 + 2x$.
practice Differentiate $f(x) = x \sin(x)$ using the product rule.
practice Differentiate $f(x) = (x+1)/(x-1)$ using the quotient rule.
practice Differentiate $f(x) = x^4 + 3 \sin(x) - 5$.
practice Differentiate $f(x) = (x^2+1) \cos(x)$ using the product rule.
practice Differentiate $f(x) = \sin(x)/x$ using the quotient rule.
practice Using the limit definition directly, find $\lim_{h\to 0} [(x+h)^3 - x^3]/h$.
JEE Using first principles, find $f'(3)$ for $f(x) = x^2$.
$0$
$7$
$6$
$3$
JEE Using first principles, find $f'(1)$ for $f(x) = x^2 + 2x$.
$2$
$5$
$0$
$4$
JEE Using first principles, find $f'(2)$ for $f(x) = x^2 - 4x$.
$0$
$-4$
$1$
$8$
JEE Given $f(1)=1$, $f'(1)=2$, $g(1)=2$, $g'(1)=1$, find $d/d x[2f(x) - 5g(x)]$ at $x=1$.
$9$
$-1$
$4$
$-8$
JEE Given $f(1)=1$, $f'(1)=2$, $g(1)=2$, $g'(1)=1$, find $d/d x[f(x) \cdot g(x)]$ at $x=1$.
$2$
$4$
$5$
$3$
JEE Given $f(1)=1$, $f'(1)=2$, $g(1)=2$, $g'(1)=1$, find $d/d x[f(x)/g(x)]$ at $x=1$.
$3/2$
$-3/4$
$1/2$
$3/4$
JEE Differentiate $f(x) = x \cdot \sin(x)$ using the product rule. Find $f'(\pi)$.
$-\pi$
$-1$
$0$
$\pi$
JEE Differentiate $f(x) = x^3 \cdot \cos(x)$ using the product rule. Find $f'(\pi/2)$.
$-3\pi^2/4$
$-\pi^3/8$
$0$
$\pi^3/8$
JEE Differentiate $f(x) = \sin(x) \cdot \cos(x)$ using the product rule. Find $f'(\pi/4)$.
$-1/2$
$1/2$
$0$
$1$
JEE Differentiate $f(x) = x^2/(x+2)$ using the quotient rule. Find $f'(1)$.
$5/3$
$-5/9$
$2/9$
$5/9$
JEE Differentiate $f(x) = (2x+1)/(x-3)$ using the quotient rule. Find $f'(1)$.
$-7/4$
$7/2$
$7/4$
$1/2$
JEE Differentiate $f(x) = (x-1)/(x+1)$ using the quotient rule. Find $f'(2)$.
$2/3$
$2/9$
$-2/9$
$1/9$
JEE Find $d/d x(x^10)$ at $x=2$.
$10240$
$4608$
$5120$
$11264$
JEE Find $d/d x(x^6)$ at $x=-1$.
$6$
$-5$
$7$
$-6$
JEE Find $d/d x(x^5)$ at $x=2$.
$80$
$160$
$64$
$192$
JEE Find $d/d x[\sin(x)]$ at $x=\pi/3$.
$\sqrt{3}/2$
$1/2$
$-1/2$
$\sqrt{3}$
JEE Find $d/d x[\cos(x)]$ at $x=\pi/6$.
$\sqrt{3}/2$
$1/2$
$-1/2$
$\sqrt{3}/3$
JEE Find $d/d x[\sin(x) - \cos(x)]$ at $x=\pi/4$.
$0$
$\sqrt{2}/2$
$-\sqrt{2}$
$\sqrt{2}$
JEE Find $d/d x[x^4 - \sin(x)]$ at $x=\pi$.
$1+4\pi^3$
$4\pi^3$
$4\pi^3-1$
$1+4\pi^4$
JEE Find $d/d x[2x^2 + 3\cos(x)]$ at $x=\pi/2$.
$2\pi$
$2\pi-3$
$2\pi+3$
$\pi^2-3$
JEE Find $d/d x[x^3 - 5\sin(x) + 2]$ at $x=0$.
$0$
$5$
$-5$
$-3$
JEE For $y=x^3$, compare the TRUE derivative $d y/d x$ at $x=2$ (from the definition) against the finite ratio $[f(3)-f(2)]/(3-2)$ a student computes and calls '$d y/d x$'. What is the true $d y/d x$ at $x=2$?
$19$
$12$
$7$
$13$
JEE For $y=x^2$, compare the TRUE derivative $d y/d x$ at $x=1$ against a finite ratio a student computes and calls '$d y/d x$'. What is the true $d y/d x$ at $x=1$?
$3$
$1$
$2$
$4$
JEE For $y=x^2$, compare the TRUE derivative $d y/d x$ at $x=5$ against a finite ratio a student computes and calls '$d y/d x$'. What is the true $d y/d x$ at $x=5$?
$9$
$11$
$12$
$10$
JEE Differentiate $x^2 \cdot \sin(x) + 3x$, choosing the right combination of rules. Find $f'(\pi)$.
$-\pi^2$
$3-2\pi$
$3+\pi^2$
$3-\pi^2$
JEE Differentiate $x^3 \cdot \sin(x) - 2x^2$, choosing the right combination of rules. Find $f'(\pi)$.
$-\pi(\pi^2+4)$
$-\pi^3$
$-\pi(3\pi+4)$
$\pi(4-\pi^2)$
JEE Differentiate $(x+1)^2 \cdot \sin(x)$, choosing the right combination of rules (expand the square first, no chain rule needed). Find $f'(0)$.
$0$
$1$
$2$
$3$
Answers
$7x^6 - 12x^3 + 2$
$\sin(x) + x \cos(x)$
$-2/(x-1)^2$
$-\sin(x)$
$-7$
$4x^3 + 3 \cos(x)$
$2x \cos(x) - (x^2+1) \sin(x)$
$(x \cos(x) - \sin(x))/x^2$
$3x^2$
$6$
$4$
$0$
$-1$
$5$
$3/4$
$-\pi$
$-\pi^3/8$
$0$
$5/9$
$-7/4$
$2/9$
$5120$
$-6$
$80$
$1/2$
$-1/2$
$\sqrt{2}$
$1+4\pi^3$
$2\pi-3$
$-5$
$12$
$2$
$10$
$3-\pi^2$
$-\pi(\pi^2+4)$
$1$
The two curves cross, which is the part worth noticing: near x = 2.5 they have the same height, and everything else about them differs. A reader who classifies by what an expression CONTAINS will put these two in the same box, because they contain the same kinds of thing. The classification that works is the second column, and it is a reading of the expression’s outermost operation, not of its ingredients.
Twenty of this set’s items end in “at x = a”, and the number they ask for is a height on the second curve, not on the first. Read the tie either way — the orange curve’s height at an x is how steep the indigo one is there, which is what the tangents check.
Both curves are drawn from the unsimplified quotient, not from the tidy expression in the table’s third column - which is why they are broken at h = 0 at all. Cancelling the h is the step that turns an expression with no value at h = 0 into one that has a value there. The picture is what that step does: it fills the hole. The fourth row is the case worth being ready for - the quotient reduces to h itself, so the answer is 0 and nothing has gone wrong.
board-easy Which of these equals $\lim_{x\to 0} (1-\cos(3x))/x$?
$0$
$3$
$1$
does not exist
board-easy Let $f(x) = x+3$ for $x<2$ and $f(x) = x^2-1$ for $x\geq 2$. Does $\lim_{x\to 2} f(x)$ exist? Justify using the left-hand and right-hand limits.
board-easy Differentiate $f(x) = 5x^4 - 2x^3 + 7$ using the power rule.
board-standard Let $f(x) = x^2-1$ for $x\leq 1$ and $f(x) = 2x-2$ for $x>1$. Does $\lim_{x\to 1} f(x)$ exist? Justify using the left-hand and right-hand limits.
board-standard Differentiate $f(x) = (3x^2+1)(2x-5)$ using the product rule.
board-standard Differentiate $f(x) = (2x-1)/(3x+4)$ using the quotient rule.
board-standard Differentiate $f(x) = x^2 \tan(x)$ using the product rule.
board-standard Differentiate $f(x) = \cos(x)/(x+1)$ using the quotient rule.
board-standard Which of these equals $d/d x[\tan(x)]$ at $x=\pi/4$?
$2$
$1$
$\sqrt{2}$
$1/2$
board-standard Using first principles, find $f'(1)$ for $f(x) = x^2 + 3x$.
board-standard Which of these equals $\lim_{x\to 0} \tan(2x)/\sin(4x)$?
$1/2$
$2$
$1$
$1/4$
JEE If $\lim_{x\to 3} (x^2+a x+b)/(x-3) = 7$, find $a+b$.
JEE Let $f(x) = 2x+3$ for $x<1$ and $f(x) = k x^2$ for $x\geq 1$. Find the value of $k$ for which $\lim_{x\to 1} f(x)$ exists.
JEE Which of these equals $\lim_{x\to 0} (1-\cos(4x))/x^2$?
$8$
$4$
$2$
$16$
JEE Using first principles, find $f'(x)$ for $f(x) = \sqrt{x}$.
JEE Differentiate $f(x) = x^2 (x+1)(2x-3)$, choosing the right combination of rules.
JEE Find the point on the curve $y = x^2-4x+3$ where the tangent is parallel to the x-axis.
JEE Which of these equals $\lim_{x\to 0} [\sqrt{1+x}-1]/x$?
$1/2$
$1$
$2$
$0$
JEE Given $f(1)=3$, $f'(1)=-2$, $g(1)=4$, $g'(1)=1$, find $d/d x[f(x)/g(x) + 3f(x)]$ at $x=1$.
Answers
$5$
$5/2$
$-14$
$6$
$4$
$0$
does not exist — LHL $=5$, RHL $=3$
$20x^3 - 6x^2$
$12x^2 + 2\cos(x)$
$4$
$1/4$
$1$
$-11/4$
$5/3$
$0$
$18x^2-30x+2$
$11/(3x+4)^2$
$2x \tan(x) + x^2 \sec^2(x)$
$-[(x+1)\sin(x)+\cos(x)]/(x+1)^2$
$2$
$5$
$1/2$
$-11$
$5$
$8$
$1/(2\sqrt{x})$
$8x^3-3x^2-6x$
$(2,-1)$
$1/2$
$-107/16$
One function, and the chapter’s two questions sitting on it at once. At x = 1 the denominator vanishes and there is nothing to substitute, so the question can only be a limit: factor, cancel, and the answer is the height of the open circle. At x = 2 the denominator survives, the function has an ordinary value, and the question can be a derivative: the slope of the line resting on the curve there. What decides which question a chapter-end problem is asking is not the wording — it is whether the denominator survives at the point named.
Twenty-two of the thirty problems are board tier, which is worth knowing before starting rather than discovering at problem twenty-three. The two board bars hold nothing that has not already been done at least once earlier in the chapter. The eight in the JEE bar are the ones where the work is deciding which rule applies, not applying it. A reader can plan a sitting off this.
Parallel to the x-axis is a statement about slope, and the chapter has already built the thing that computes one. The level tangent holds the same one-unit gap from the axis at both ends — that is what parallel means — while every other tangent tilts. So set 2x - 4 = 0, and the point is (2, -1).
In computing $f'(1) = \lim_{h → 0} [(1+h)^2-1]/h$ for $f(x)=x^2$, the key algebraic step is simplifying $(1+h)^2-1$ before dividing by $h$. What does this simplify to? Why does that matter?
$(1+h)^2-1 = h^2-1$, which cannot be simplified further. So the limit must be found by trying values of $h$ close to 0.
$(1+h)^2-1 = 2h$. Dividing by h gives exactly 2 for every value of h, not just as h approaches 0.
$(1+h)^2-1 = 1+h^2$, which is never zero. So the original expression is already defined at $h=0$ without needing a limit at all. That works the same way a polynomial’s value can simply be read off by direct substitution.
$(1+h)^2-1 = 2h+h^2 = h(2+h)$. Factoring out $h$ lets it cancel with the denominator, leaving $2+h$, which is defined even at $h=0$ and gives the limit 2 directly.
Check your answer
$(1+h)^2-1 = h^2-1$, which cannot be simplified further. So the limit must be found by trying values of $h$ close to 0. — $(1+h)^2$ expands to $1+2h+h^2$, not $h^2$ — the $2h$ cross term is easy to lose. Subtracting 1 gives $2h+h^2$, which DOES simplify further (factor out $h$).
$(1+h)^2-1 = 2h$. Dividing by h gives exactly 2 for every value of h, not just as h approaches 0. — $2h/h = 2$ only after cancelling — but $(1+h)^2-1$ is $2h+h^2$, not $2h$ alone. The $h^2$ term survives division as an extra $+h$, which is exactly why the LIMIT (not a fixed value) matters: $(2+h) → 2$ only as $h → 0$.
$(1+h)^2-1 = 1+h^2$, which is never zero. So the original expression is already defined at $h=0$ without needing a limit at all. That works the same way a polynomial’s value can simply be read off by direct substitution. — $(1+h)^2 = 1+2h+h^2$ — the middle term $2h$ cannot be dropped. Once it’s included, $(1+h)^2-1 = 2h+h^2$ is 0 at $h=0$, which is exactly the $0/0$ shape a limit is needed to resolve — not a sign that no limit is needed.
✓ $(1+h)^2-1 = 2h+h^2 = h(2+h)$. Factoring out $h$ lets it cancel with the denominator, leaving $2+h$, which is defined even at $h=0$ and gives the limit 2 directly. — (D) Factoring $(1+h)^2-1$ as $h(2+h)$ is what lets the $h$ in the denominator cancel — the resulting $2+h$ is perfectly well-defined at $h=0$, unlike the original $0/0$-shaped expression.
The zero-over-zero technique proves the general result $\lim_{x → a} (x^n-a^n)/(x-a) = n \cdot a^{n-1}$ for any positive integer $n$. What is the key factoring identity that makes this work?
$x^n-a^n = (x-a)(x^{n-1}+x^{n-2}a+\ldots+a^{n-1})$. The second factor has exactly $n$ terms, and cancelling $(x-a)$ leaves that sum. That sum equals $n \cdot a^{n-1}$ once every $x$ is replaced by $a$.
$x^n-a^n = (x-a)^n$. So cancelling one factor of $(x-a)$ leaves $(x-a)^{n-1}$, which is 0 at $x=a$. That matches the pattern of raising a whole binomial to the same power $n$ that already appears on $x$ and $a$ separately.
$x^n-a^n = n(x-a)$ directly, so the limit is just $n$, with no dependence on $a$.
$x^n-a^n$ cannot be factored in general; the formula only works when $n=2$ or $n=3$.
Check your answer
✓ $x^n-a^n = (x-a)(x^{n-1}+x^{n-2}a+\ldots+a^{n-1})$. The second factor has exactly $n$ terms, and cancelling $(x-a)$ leaves that sum. That sum equals $n \cdot a^{n-1}$ once every $x$ is replaced by $a$. — (A) The $n$-term factor is what turns into $n \cdot a^{n-1}$ once $(x-a)$ cancels and $x$ is replaced by $a$ — this is the general engine behind the power rule.
$x^n-a^n = (x-a)^n$. So cancelling one factor of $(x-a)$ leaves $(x-a)^{n-1}$, which is 0 at $x=a$. That matches the pattern of raising a whole binomial to the same power $n$ that already appears on $x$ and $a$ separately. — $(x-a)^n$ means $(x-a)$ multiplied by itself $n$ times — a completely different expression from $x^n-a^n$. The real identity factors $x^n-a^n$ into $(x-a)$ times a SUM of $n$ terms, not a power of $(x-a)$.
$x^n-a^n = n(x-a)$ directly, so the limit is just $n$, with no dependence on $a$. — $x^n-a^n$ is not proportional to $(x-a)$ alone — try $n=2$: $x^2-a^2 = (x-a)(x+a)$, not $n(x-a)=2(x-a)$. The correct factor is a sum of $n$ terms, which is why the final answer depends on $a$ as $n \cdot a^{n-1}$, not just on $n$.
$x^n-a^n$ cannot be factored in general; the formula only works when $n=2$ or $n=3$. — The identity $x^n-a^n = (x-a)(x^{n-1}+x^{n-2}a+\ldots+a^{n-1})$ holds for ANY positive integer $n$ — it is not special to $n=2$ or $n=3$, which is exactly why this technique proves the GENERAL result used later for the power rule.
The sandwich proof compares the areas of two triangles and a sector to get sin(x) < x < tan(x) for 0 < x < $\pi/2.$ What is the very next algebraic move that turns this into the sandwich for sin(x)/x?
Divide every part by x directly, giving sin(x)/x < 1 < tan(x)/x. This already sandwiches sin(x)/x between two known limits, since both 1 and tan(x)/x look like they should approach familiar values as x shrinks toward 0.
Divide every part of sin(x) < x < tan(x) by sin(x) (positive in this range, so the inequalities keep their direction). Then take reciprocals (which flips the direction again), giving cos(x) < sin(x)/x < 1.
Take the square root of each part of sin(x) < x < tan(x) to remove the trig functions before dividing.
Multiply every part by cos(x) first, then divide by x, since tan(x) = sin(x)/cos(x).
Check your answer
Divide every part by x directly, giving sin(x)/x < 1 < tan(x)/x. This already sandwiches sin(x)/x between two known limits, since both 1 and tan(x)/x look like they should approach familiar values as x shrinks toward 0. — Dividing sin(x) < x < tan(x) by x gives sin(x)/x < 1 < tan(x)/x — three separate pieces, not a chain with sin(x)/x sandwiched between two things both approaching 1. Dividing by sin(x) instead is what produces cos(x) < sin(x)/x < 1, the actual sandwich the proof uses.
✓ Divide every part of sin(x) < x < tan(x) by sin(x) (positive in this range, so the inequalities keep their direction). Then take reciprocals (which flips the direction again), giving cos(x) < sin(x)/x < 1. — (B) Dividing by sin(x), then taking reciprocals, is exactly what produces cos(x) < sin(x)/x < 1 — the sandwich that forces sin(x)/x → 1 as cos(x) → 1.
Take the square root of each part of sin(x) < x < tan(x) to remove the trig functions before dividing. — Taking a square root would need every part of the inequality to be non-negative and would change what is being compared — it plays no role in the real derivation, which stays a simple two-step divide-then-reciprocate.
Multiply every part by cos(x) first, then divide by x, since tan(x) = sin(x)/cos(x). — Multiplying by cos(x) first changes tan(x) back into sin(x)/cos(x) . cos(x) = sin(x), which just undoes the tangent — it does not lead to the cos(x) < sin(x)/x < 1 result the atom’s own proof reaches.
The derivation of $d/d x \sin(x)$ rewrites $\sin(x+h)-\sin(x)$ using an identity. That identity is $\sin(A)-\sin(B) = 2\cos((A+B)/2)\sin((A-B)/2)$, with $A=x+h$, $B=x$. What does this substitution actually achieve?
It turns a DIFFERENCE of two sine values into a PRODUCT of a cosine term and a sine-over-half-argument term. Direct substitution just makes the original difference 0/0-shaped as h → 0. Only the second factor is the 0/0-type piece that needs the standard trig limit. The first factor, cos(x+h/2), is already well-behaved and just substitutes to cos(x).
It cancels the h entirely before any limit is taken. So the final answer cos(x) can be read off without needing a limit step at all. That is the same way the h in the numerator and the h in the denominator of an ordinary difference quotient sometimes cancel outright. This can happen before any limit is even considered. What is left behind is nothing but a plain algebraic expression, at the very end.
It converts the sine difference into a difference of two cosines instead, which is easier to evaluate directly.
It replaces the limit definition with a completely different formula that does not involve h at all.
Check your answer
✓ It turns a DIFFERENCE of two sine values into a PRODUCT of a cosine term and a sine-over-half-argument term. Direct substitution just makes the original difference 0/0-shaped as h → 0. Only the second factor is the 0/0-type piece that needs the standard trig limit. The first factor, cos(x+h/2), is already well-behaved and just substitutes to cos(x). — (A) The identity splits one hard 0/0-shaped difference into a well-behaved factor (cos(x+h/2)) and one factor that IS the standard trig limit in disguise (sin(h/2)/(h/2)) — isolating exactly where the limit work is still needed.
It cancels the h entirely before any limit is taken. So the final answer cos(x) can be read off without needing a limit step at all. That is the same way the h in the numerator and the h in the denominator of an ordinary difference quotient sometimes cancel outright. This can happen before any limit is even considered. What is left behind is nothing but a plain algebraic expression, at the very end. — The identity turns the difference into a PRODUCT, one factor of which (sin(h/2)/(h/2)) is still a 0/0-shaped limit as h → 0. The limit step is not skipped — it is relocated to exactly that one factor.
It converts the sine difference into a difference of two cosines instead, which is easier to evaluate directly. — $\sin(A)-\sin(B) = 2\cos((A+B)/2)\sin((A-B)/2)$ is a product of a cosine term and a sine term, not a difference of two cosines — that is a different identity ($\cos(A)-\cos(B)$) used for the companion result $d/d x \cos(x)=-\sin(x)$.
It replaces the limit definition with a completely different formula that does not involve h at all. — h still appears twice in cos(x+h/2).[sin(h/2)/(h/2)] — once inside the cosine, once inside the sine-over-half-argument factor. The identity reshapes the expression; it does not remove h from it.
The algebra-of-limits laws all share one requirement before they can be used. What is it? Why does it matter?
At least one of f or g must be a polynomial. The laws only work when direct substitution is already possible. That is the same restriction that would govern exactly when a limit could be read off a formula without any special technique at all.
Both $\lim_{x → a} f(x)$ and $\lim_{x → a} g(x)$ must already be known to exist. The laws combine two existing limits into a new one. They cannot manufacture a limit for a function whose own limit is unknown or does not exist.
f and g must be equal to each other at x=a — the laws only combine limits of identical functions.
The point $a$ must be inside the domain of both $f$ and $g$. The laws require $f(a)$ and $g(a)$ to be defined.
Check your answer
At least one of f or g must be a polynomial. The laws only work when direct substitution is already possible. That is the same restriction that would govern exactly when a limit could be read off a formula without any special technique at all. — The laws apply to any f and g whose limits already exist — trigonometric, rational, or otherwise — not only polynomials. What matters is that l and m are already known, not what kind of function produced them.
✓ Both $\lim_{x → a} f(x)$ and $\lim_{x → a} g(x)$ must already be known to exist. The laws combine two existing limits into a new one. They cannot manufacture a limit for a function whose own limit is unknown or does not exist. — (B) The laws are a combining tool, not a manufacturing tool — both input limits must already be established before a law can produce a new one.
f and g must be equal to each other at x=a — the laws only combine limits of identical functions. — The whole point of the laws is combining two DIFFERENT functions’ limits into a new one (like sums or products of unrelated quantities) — requiring f and g to be identical would make the laws pointless.
The point $a$ must be inside the domain of both $f$ and $g$. The laws require $f(a)$ and $g(a)$ to be defined. — This reverses the actual dependency: the laws need lim $f(x)$ and lim $g(x)$ to exist, which says nothing about whether $f(a)$ or $g(a)$ are defined. A limit can exist at a point neither function is even defined at.
Why does direct substitution always work for a limit of a polynomial? Why does it only SOMETIMES work for a limit of a rational function $f(x)/g(x)$?
Polynomials are always continuous everywhere. Rational functions are never continuous anywhere. That is the same absolute split that would mean a rational function like 1/x could never behave predictably at any single point of its domain. This would be true even where its denominator is perfectly nonzero.
Direct substitution works for any function as long as x approaches a positive number.
A polynomial is built only from sums, products and scalar multiples of $x$. The algebra of limits carries substitution through every one of those steps. It does so without exception. A rational function adds division. The quotient law only permits that division when $g(a) \neq 0$.
Rational functions require a completely different definition of limit than polynomials do.
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Polynomials are always continuous everywhere. Rational functions are never continuous anywhere. That is the same absolute split that would mean a rational function like 1/x could never behave predictably at any single point of its domain. This would be true even where its denominator is perfectly nonzero. — A rational function IS continuous at every point where its denominator is nonzero — ‘never continuous anywhere’ overstates the real, narrower exception (only at the zeros of g).
Direct substitution works for any function as long as x approaches a positive number. — Whether $a$ is positive, negative, or zero plays no role here — what matters is only whether $g(a)$ equals zero, not the sign of $a$ itself.
✓ A polynomial is built only from sums, products and scalar multiples of $x$. The algebra of limits carries substitution through every one of those steps. It does so without exception. A rational function adds division. The quotient law only permits that division when $g(a) \neq 0$. — (C) Polynomials only ever need the sum/product/scalar laws, which have no proviso. Rational functions add the quotient law, which requires $g(a) \neq 0$ — the one place substitution can fail.
Rational functions require a completely different definition of limit than polynomials do. — The limit definition — what $f(x)$ approaches as $x$ approaches $a$ — is identical for both. What differs is which ALGEBRA-OF-LIMITS law applies: polynomials only need sum/product/scalar laws; rationals also need the quotient law, which carries its own proviso.
The sandwich proof for $\lim_{x → 0} \sin(x)/x=1$ is built for $0 < x < \pi/2$ only. Why does the SAME result still hold as $x$ approaches 0 from the negative side?
sin(x)/x is an even function of x (sin(−x)/(−x) = sin(x)/x). So the value it approaches from the left is identical to the value it approaches from the right. The one-sided proof for x>0 covers both sides for free.
The proof also works for x<0 because sin(x) and x are both negative there. So the areas compared in the proof stay positive automatically.
It does not actually hold from the left. The two one-sided limits differ, so the two-sided limit technically does not exist. That is the same conclusion a student would correctly reach for a function with a genuine jump at that exact point.
Because sin(x)/x is a polynomial for small x, and polynomials are symmetric near the origin.
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✓ sin(x)/x is an even function of x (sin(−x)/(−x) = sin(x)/x). So the value it approaches from the left is identical to the value it approaches from the right. The one-sided proof for x>0 covers both sides for free. — (A) sin(x)/x is even — its value is identical whether x approaches 0 from above or below — so proving the result for x>0 automatically settles the x<0 side too.
The proof also works for x<0 because sin(x) and x are both negative there. So the areas compared in the proof stay positive automatically. — The triangle/sector area comparison sin(x) < x < tan(x) is set up specifically for 0 < x < $\pi/2$ and is not separately re-derived for negative x — the atom’s own claim explicitly names evenness, not a re-run of the areas argument, as the reason the negative side is covered.
It does not actually hold from the left. The two one-sided limits differ, so the two-sided limit technically does not exist. That is the same conclusion a student would correctly reach for a function with a genuine jump at that exact point. — The atom’s claim is explicit that the identical result holds from the left, via evenness — LHL and RHL do agree at 1. Claiming they differ contradicts the standard result directly.
Because sin(x)/x is a polynomial for small x, and polynomials are symmetric near the origin. — sin(x)/x is a genuinely trigonometric expression, not a polynomial, at every x including near 0 — ‘polynomials are symmetric near the origin’ is not why this particular even-function symmetry holds.
When $f(a)=0$ and $g(a)=0$ in a rational limit, $(x-a)$ can always be factored out of BOTH the numerator and denominator. Why does that make sense?
Because $f(a)=0$ and $g(a)=0$, the factor theorem guarantees that $(x-a)$ is a factor of both $f$ and $g$. A polynomial that vanishes at $x=a$ always has $(x-a)$ as one of its factors. So it can always be pulled out and cancelled.
Because every rational function can be simplified by cancelling any common-looking term from the top and bottom. This holds whether or not it is actually a shared factor. That is the same loose habit that would let a student cancel an x appearing in both places, without ever checking it is a genuine factor.
Because 0/0 is defined to equal 1 by convention, so the factoring step is just a formality.
Because f and g must be the exact same polynomial whenever they both vanish at x=a.
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✓ Because $f(a)=0$ and $g(a)=0$, the factor theorem guarantees that $(x-a)$ is a factor of both $f$ and $g$. A polynomial that vanishes at $x=a$ always has $(x-a)$ as one of its factors. So it can always be pulled out and cancelled. — (A) The factor theorem is the real guarantee: vanishing at x=a is exactly what forces (x−a) to be a genuine factor of a polynomial, licensing the cancellation.
Because every rational function can be simplified by cancelling any common-looking term from the top and bottom. This holds whether or not it is actually a shared factor. That is the same loose habit that would let a student cancel an x appearing in both places, without ever checking it is a genuine factor. — Cancellation is only valid for a genuinely shared FACTOR — something the factor theorem actually guarantees exists here. Cancelling terms that merely look alike, without that guarantee, is not valid algebra.
Because 0/0 is defined to equal 1 by convention, so the factoring step is just a formality. — 0/0 stays undefined throughout — factoring never assigns it a value. The technique avoids ever evaluating the 0/0 form directly, by cancelling (x-a) BEFORE substituting x=a.
Because f and g must be the exact same polynomial whenever they both vanish at x=a. — f and g only need to share the single root x=a — they can be completely different polynomials otherwise. The factor theorem guarantees (x-a) divides each of them separately, not that f and g are the same.
The derivation of $d/d x \sin(x) = \cos(x)$ rewrites the difference quotient as $\cos(x+h/2) \cdot [\sin(h/2)/(h/2)]$. Why is the standard trig limit $\lim_{t → 0} \sin(t)/t=1$ essential at this exact step?
Without it, cos(x+h/2) could not be evaluated at all, since cosine has no known value near x.
As h → 0, the bracket sin(h/2)/(h/2) is itself a 0/0-type limit. Direct substitution gives 0/0, undefined. Only the standard trig limit result resolves it to exactly 1, leaving cos(x+h/2) → cos(x) as the whole answer.
It converts the sine difference quotient into a polynomial, which can then be handled by the algebra of limits alone.
It is only a simplifying convenience. The same derivative could be reached just as easily without it, using only the algebra of limits. That works the same way a shortcut sometimes speeds up a calculation, without ever being strictly necessary to reach the final answer.
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Without it, cos(x+h/2) could not be evaluated at all, since cosine has no known value near x. — cos(x+h/2) needs no special limit result — as h → 0 it just substitutes directly to cos(x), the same way any continuous function would. The bracket sin(h/2)/(h/2) is the piece that is 0/0-shaped and needs the standard result.
✓ As h → 0, the bracket sin(h/2)/(h/2) is itself a 0/0-type limit. Direct substitution gives 0/0, undefined. Only the standard trig limit result resolves it to exactly 1, leaving cos(x+h/2) → cos(x) as the whole answer. — (B) The bracket sin(h/2)/(h/2) is a 0/0-shaped limit in its own right — exactly the case the standard trig limit result exists to resolve; without it, this one factor is stuck.
It converts the sine difference quotient into a polynomial, which can then be handled by the algebra of limits alone. — The final answer cos(x) is still a trig function, not a polynomial — nothing about this derivation converts sine or cosine into polynomial form at any stage.
It is only a simplifying convenience. The same derivative could be reached just as easily without it, using only the algebra of limits. That works the same way a shortcut sometimes speeds up a calculation, without ever being strictly necessary to reach the final answer. — Without the standard trig limit, sin(h/2)/(h/2) is exactly a 0/0-type limit — direct substitution gives 0/0, and the algebra-of-limits laws (which require both input limits to already exist) cannot resolve an indeterminate form on their own. That is precisely why sin(x)/x → 1 needed its own separate sandwich proof in the first place.
In computing $f'(1) = \lim_{h → 0} [(1+h)^2-1]/h$ for $f(x)=x^2$, the key algebraic step is simplifying $(1+h)^2-1$ before dividing by $h$. What does this simplify to? Why does that matter?
$(1+h)^2-1 = h^2-1$, which cannot be simplified further. So the limit must be found by trying values of $h$ close to 0.
$(1+h)^2-1 = 2h$. Dividing by h gives exactly 2 for every value of h, not just as h approaches 0.
$(1+h)^2-1 = 1+h^2$, which is never zero. So the original expression is already defined at $h=0$ without needing a limit at all. That works the same way a polynomial’s value can simply be read off by direct substitution.
$(1+h)^2-1 = 2h+h^2 = h(2+h)$. Factoring out $h$ lets it cancel with the denominator, leaving $2+h$, which is defined even at $h=0$ and gives the limit 2 directly.
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$(1+h)^2-1 = h^2-1$, which cannot be simplified further. So the limit must be found by trying values of $h$ close to 0. — $(1+h)^2$ expands to $1+2h+h^2$, not $h^2$ — the $2h$ cross term is easy to lose. Subtracting 1 gives $2h+h^2$, which DOES simplify further (factor out $h$).
$(1+h)^2-1 = 2h$. Dividing by h gives exactly 2 for every value of h, not just as h approaches 0. — $2h/h = 2$ only after cancelling — but $(1+h)^2-1$ is $2h+h^2$, not $2h$ alone. The $h^2$ term survives division as an $\text{extra} +h$, which is exactly why the LIMIT (not a fixed value) matters: $(2+h) → 2$ only as $h → 0$.
$(1+h)^2-1 = 1+h^2$, which is never zero. So the original expression is already defined at $h=0$ without needing a limit at all. That works the same way a polynomial’s value can simply be read off by direct substitution. — $(1+h)^2 = 1+2h+h^2$ — the middle term $2h$ cannot be dropped. Once it’s included, $(1+h)^2-1 = 2h+h^2$ is 0 at $h=0$, which is exactly the $0/0$ shape a limit is needed to resolve — not a sign that no limit is needed.
✓ $(1+h)^2-1 = 2h+h^2 = h(2+h)$. Factoring out $h$ lets it cancel with the denominator, leaving $2+h$, which is defined even at $h=0$ and gives the limit 2 directly. — (D) Factoring $(1+h)^2-1$ as $h(2+h)$ is what lets the $h$ in the denominator cancel — the resulting $2+h$ is perfectly well-defined at $h=0$, unlike the original $0/0$-shaped expression.
Why does direct substitution always work for a limit of a polynomial? Why does it only SOMETIMES work for a limit of a rational function $f(x)/g(x)$?
Polynomials are always continuous everywhere. Rational functions are never continuous anywhere. That is the same absolute split that would mean a rational function like 1/x could never behave predictably at any single point of its domain. This would be true even where its denominator is perfectly nonzero.
Direct substitution works for any function as long as x approaches a positive number.
A polynomial is built only from sums, products and scalar multiples of $x$. The algebra of limits carries substitution through every one of those steps. It does so without exception. A rational function adds division. The quotient law only permits that division when $g(a) \neq 0$.
Rational functions require a completely different definition of limit than polynomials do.
Check your answer
Polynomials are always continuous everywhere. Rational functions are never continuous anywhere. That is the same absolute split that would mean a rational function like 1/x could never behave predictably at any single point of its domain. This would be true even where its denominator is perfectly nonzero. — A rational function IS continuous at every point where its denominator is nonzero — ‘never continuous anywhere’ overstates the real, narrower exception (only at the zeros of g).
Direct substitution works for any function as long as x approaches a positive number. — Whether $a$ is positive, negative, or zero plays no role here — what matters is only whether $g(a)$ equals zero, not the sign of $a$ itself.
✓ A polynomial is built only from sums, products and scalar multiples of $x$. The algebra of limits carries substitution through every one of those steps. It does so without exception. A rational function adds division. The quotient law only permits that division when $g(a) \neq 0$. — (C) Polynomials only ever need the sum/product/scalar laws, which have no proviso. Rational functions add the quotient law, which requires $g(a) \neq 0$ — the one place substitution can fail.
Rational functions require a completely different definition of limit than polynomials do. — The limit definition — what $f(x)$ approaches as $x$ approaches $a$ — is identical for both. What differs is which ALGEBRA-OF-LIMITS law applies: polynomials only need sum/product/scalar laws; rationals also need the quotient law, which carries its own proviso.