Limits and continuity

THE ONE IDEA UNDER EVERYTHING

$\underset{x\to a}{\lim }f\left(x\right)=L$ means: as $x$ gets arbitrarily close to $a$ (from both sides), $f\left(x\right)$ gets arbitrarily close to $L$. The limit is about the approach, never the arrival.

Read that definition once more and notice what it never mentions: $f\left(a\right)$. The value the function takes at $a$ is not part of the question being asked. It can equal $L$, it can be some other number entirely, or the function can fail to be defined there at all — and the limit is $L$ in every one of those cases, because in every one of them the approach did exactly the same thing.

Those three pictures are the core of the unit in miniature. Every question about continuity at a particular point — does the limit exist there, is the function continuous there, what kind of discontinuity is this — is a question about how the second and third pictures differ from the first. Most limit mistakes are arrival answers to approach questions.

Not everything in the unit is that shape, and it is worth knowing which questions are not, so you don’t go looking for a single bad point when there isn’t one. A function can be broken at two points at once (Problem 7), and two of this unit’s questions are not about a single point at all: end behavior asks what happens infinitely far out (Problem 6), and the Intermediate Value Theorem makes a claim about a whole interval (Problem 8). Rung 5 and rung 7 exist for exactly those.

THE LIMIT LAWS — AND THE ONE HYPOTHESIS THAT EVER FAILS

Limits get along with arithmetic. If $\underset{x\to a}{\lim }f\left(x\right)=A$ and $\underset{x\to a}{\lim }g\left(x\right)=B$, and both of those are real numbers, then

These are the reason substitution works at all. A polynomial is built out of constants and $x$ by adding and multiplying, and the laws say a limit walks straight through both operations — so $\underset{x\to a}{\lim }$ of a polynomial is the polynomial evaluated at $a$, and every “just plug it in” you have ever done was these six lines quietly doing their job.

The quotient law is the one with a condition, and that condition is the entire unit. $B\ne 0$ is not fine print. When the bottom limit is zero, the law does not apply — and that is exactly the moment substitution hands you $\frac{\text{nonzero}}{0}$ or $\frac{0}{0}$ and the decision tree takes over. So the awkward cases in this guide are not places where the rules broke. They are places where a rule politely declined, and the rest of the packet is what you do instead.

Reading them backwards is a trap the exam likes. The laws run one direction only: if the pieces have limits, the combination does. Not the reverse. Both $\underset{x\to 0}{\lim }\frac{\left|x\right|}{x}$ and $\underset{x\to 0}{\lim }\left(-\frac{\left|x\right|}{x}\right)$ fail to exist, and their sum is $0$ everywhere except the origin, so the sum’s limit is a perfectly good $0$. “$\lim (f+g)$ exists” tells you nothing about $f$ and $g$ on their own.

Where this shows up. Most often you are handed $A$ and $B$ — by a table, by a graph, or in a sentence — and asked to assemble something from them without ever seeing a formula. If $\underset{x\to 2}{\lim }f\left(x\right)=3$ and $\underset{x\to 2}{\lim }g\left(x\right)=-5$, then

and the only judgment involved was checking that the denominator’s limit wasn’t zero before using the quotient law. Say that check out loud; it is the part worth a point.

SUBSTITUTION IS A DIAGNOSIS, NOT AN ATTEMPT

Once rungs 1 and 2 are clear — you have a formula rather than a picture, and one formula rather than a piecewise rule with a seam at $a$ — every limit in this guide starts the same way: put $a$ in and see what comes back. That is not a guess, and it is not a first try that might fail. It is a test, and its result names your next move.

Why substitution works when it works. Plugging in is legal exactly when the function is continuous at $a$ — and polynomials are continuous everywhere, while a rational function is continuous everywhere its denominator isn’t zero. So for most of what you meet in this unit, “I can substitute” and “the function is continuous here” are the same sentence said two different ways. That is why the continuity material later in this guide is not a new topic bolted on at the end: it is the rule that made step one legal in the first place.

It is also the reason the seam question sits above this one on the ladder. A piecewise function has no single formula at its seam, so there is nothing there for “continuous at $a$” to be true of yet — and substituting anyway will hand you a number from whichever piece owns the endpoint, with no warning that the other piece went somewhere else entirely. Problem 5(c) is that exact number.

Outcome 1 — a number. You are done. Looking for something cleverer is how students turn a finished problem into a wrong one. If substitution returned $\frac{9}{6}$, the answer is $\frac{3}{2}$, and the fact that it felt too easy is not evidence of anything.

Outcome 2 — $\frac{\text{nonzero}}{0}$. The denominator is collapsing toward zero while the numerator is not. So you are dividing a number that stays put by a number that becomes arbitrarily small, and the quotient grows without bound. No algebra fixes this, because there is nothing to cancel — that zero is real and it isn’t going anywhere. Your remaining job is a sign question: which direction does it blow up on each side?

Outcome 3 — $\frac{0}{0}$. Top and bottom are both collapsing, and the answer depends on which of them collapses faster — which is precisely the thing the symbol $\frac{0}{0}$ cannot tell you. This is called an indeterminate form, and the phrase means what it says: not “undefined,” but “not yet determined.” Compare $\underset{x\to 0}{\lim }\frac{x}{x}=1$, $\underset{x\to 0}{\lim }\frac{2x}{x}=2$, and $\underset{x\to 0}{\lim }\frac{x^2}{x}=0$ — all three substitute to $\frac{0}{0}$, and all three have different answers. The form is a question, not a result.

For the algebraic limits in this guide, both top and bottom hitting zero at $x=a$ means both of them contain the factor $(x-a)$. Finding it is the whole task.

THE $\frac{0}{0}$ PLAYBOOK

$\frac{0}{0}$ is an instruction to go looking. In the three algebraic shapes below the instruction is the same in substance — expose the shared factor and remove it — and only the technique changes, depending on how the factor is hiding. The fourth row is the exception worth knowing about before it turns up on a test: when the collapsing pieces are trigonometric, there is no shared factor to find, and no amount of algebra will produce one.

Why canceling is legal. This is the step students feel guilty about, so it is worth being exact. When you write

you have not divided by zero. The limit only ever looks at $x$ values near $a$ and never at $a$ itself — that is what “as $x$ approaches $a$” means. For every one of those $x$, the quantity $x-a$ is a genuine nonzero number, and dividing by a nonzero number is something you have been allowed to do since seventh grade. The two functions above disagree at exactly one point, $x=a$, and a limit cannot tell them apart, because a limit never asks about that point.

The shape the problems in this guide don’t show you. Complex fractions turn up on the exam and nowhere else in this packet, so here is the move in full. Suppose

Substituting gives $\frac{0}{0}$, and there is nothing to factor and no root to conjugate — because the numerator isn’t one object yet. Combine it over a common denominator first:

so the whole expression becomes

using $2-x=-(x-2)$ to make the shared factor visible. Now substitute again: $\frac{-1}{2\left(2\right)}=-\frac{1}{4}$. The pattern is the same one you have been running all along — the combining step existed only to produce something that could be canceled.

THE $\frac{0}{0}$ THAT ISN’T ALGEBRA: SQUEEZE AND THE TRIG LIMITS

Every other $\frac{0}{0}$ in this guide had a factor hiding in it. A trigonometric one does not. There is no $x$ tucked inside $\sin x$ waiting to be pulled out, so factoring, conjugates, and combining all have nothing to grip — and a student who has only ever been told “$\frac{0}{0}$ means do algebra” will grind at $\frac{\sin x}{x}$ until the clock runs out. Two results replace the algebra. Both are worth memorizing outright:

The first says something you can see: very near zero, $\sin x$ and $x$ are the same number to as many decimals as you like. ($\sin \left(0.01\right)=0.00999983\dots$) The graph of $\sin x$ leaves the origin along the line $y=x$ and only later peels away, so the ratio of the two starts at $1$. That is also why the limit needs $x$ in radians — in degrees the sine curve leaves the origin along a much flatter line and the ratio is $\frac{\pi }{180}$ instead.

Matching the pattern. The result is not really about $x$. It says $\frac{\sin u}{u}\to 1$ for any quantity $u$ that is going to zero — the angle inside the sine and the thing underneath just have to be the same. So make them match, and pay for it with a constant:

Multiplying top and bottom by $5$ costs nothing and buys the pattern. That one move handles nearly every trig limit the exam asks for.

The squeeze theorem. When a function is too wild to evaluate but you can trap it, trap it. If

then $\underset{x\to a}{\lim }f\left(x\right)=L$ as well. The argument is not subtle: the two outer functions are closing in on the same place and $f$ is pinned between them, so it has nowhere else to go.

The trap to set is almost always the same one: a sine or a cosine is stuck between $-1$ and $1$ no matter what you feed it, so multiply that sandwich by whatever is out front and you have your $g$ and $h$. And note what the squeeze does not require — it never asks you to evaluate $f$, or even to know what $f$ is doing. That is the whole point of using it.

ONE-SIDED LIMITS AND EXISTENCE

$\underset{x\to a^{-}}{\lim }$ means: approach $a$ from the left, using only $x$ values less than $a$. $\underset{x\to a^{+}}{\lim }$ uses only $x$ values greater than $a$. The two-sided limit exists only when both of these arrive at the same place — and “a place” has to mean an actual number:

That word finite is doing real work, and leaving it out is a mistake you can make while doing everything else right. Take $\frac{1}{x^2}$ at $x=0$: both sides run to $+\infty$, so in a loose sense they “agree.” But $+\infty$ is not a number the function is getting arbitrarily close to — it is shorthand for the function escaping every number you name. The two sides matching there does not rescue the limit; nothing does. Agreement is necessary, not sufficient.

Notice what the picture does not show: anything wrong with either one-sided limit. Both exist, both are perfectly ordinary numbers. The failure is entirely in the comparison. That is why the honest way to report a nonexistent limit is to name both sides and then say they disagree — “does not exist” on its own throws away everything you actually found out.

When a limit fails to exist, there are only three ways it can happen, and saying which is part of the answer:

• the two sides disagree — a jump;
• the function is unbounded — it runs off to $\pm \infty$;
• the function oscillates without settling, like $\sin (1∕x)$ near $0$, which crosses every value between $-1$ and $1$ infinitely often no matter how close you get.

INFINITE LIMITS: WHEN THE DENOMINATOR ALONE GOES TO ZERO

This is outcome 2 from the diagnosis — $\frac{\text{nonzero}}{0}$ — and it means the graph has a vertical asymptote at $x=a$. Writing $\underset{x\to a}{\lim }f\left(x\right)=+\infty$ is not claiming the limit equals some number called infinity. It is a description of how the limit fails to exist: the function isn’t wandering or jumping, it is climbing without bound. That description is worth more credit than “DNE,” so give it.

How to find the sign on each side. You are not guessing. Look at the factor in the denominator that is producing the zero, and ask what sign it has just to the left of $a$ and just to the right. If the fraction had a common factor and you canceled it first, read the power that is left in the denominator, not the one you started with — Problem 7 is exactly that case.

If counting powers feels abstract, there is a fallback that never fails: pick a number just to the left of $a$ and one just to the right, and read the signs of the top and bottom separately instead of computing the value. Take $\frac{x-2}{x-4}$ near $x=4$. At $x=3.9$ the top is $1.9$ — positive — and the bottom is $-0.1$ — negative and tiny — so the quotient is large and negative: the function runs to $-\infty$ from the left. At $x=4.1$ the top is still positive and the bottom is $+0.1$, so the quotient is large and positive: $+\infty$ from the right. Write those two checks down. Doing them in your head is where this goes wrong.

LIMITS AT INFINITY: END BEHAVIOR AND THE $\left|x\right|$ TRAP

$x\to \pm \infty$ is a different question wearing the same symbol. Here nothing is dividing by zero and nothing is blowing up at a point — you are asking what the function settles into far out along the axis. A finite answer $L$ means a horizontal asymptote $y=L$.

Dominant terms. Out at large $\left|x\right|$, the highest power in a polynomial is so much bigger than everything else that the rest are rounding error. In $5x^3-2x$ at $x=1000$, the first term is $5\times 10^9$ and the second is $2000$: the second term is not small, it is irrelevant. So for a rational function, compare the top and bottom degrees:

• top degree $<$ bottom degree $\Rightarrow$ the limit is $0$ (the denominator outruns the numerator);
• degrees equal $\Rightarrow$ the ratio of the leading coefficients (they grow at the same rate, so the ratio settles);
• top degree $>$ bottom degree $\Rightarrow \pm \infty$ (the numerator outruns the denominator — and now you must decide which sign).

Why that shortcut is legitimate. It is the result of dividing top and bottom by the highest power in the denominator. For example,

because every term of the form $\frac{c}{x^n}$ goes to $0$. The “circle the dominant term” rule is just this calculation with the zeroes already taken. If you ever doubt the shortcut on an unusual problem, do the division — it always works and it is never wrong.

The root trap. $\sqrt{x^2}=\left|x\right|$, not $x$. The square root symbol returns the nonnegative root, always, so for $x<0$ you must write $\sqrt{x^2}=-x$ — which is a positive number, because $x$ itself is negative. Concretely, $\sqrt{9x^2}=3\left|x\right|$, and

This is the single most-missed step in the unit, and its consequence is that a function with a square root on top can have two different horizontal asymptotes — one as $x\to +\infty$ and a different one as $x\to -\infty$. Problem 6 draws that picture.

CONTINUITY AT A POINT: THE THREE-PART CHECKLIST

Informally, $f$ is continuous at $a$ if you can draw through the point without lifting your pencil. Formally — and this is the version that scores — all three of these must hold:

Run them in that order. Each one is checking something the previous one assumed: (1) asks whether there is an arrival at all, (2) asks whether there is an approach at all, and (3) asks whether the two agree. You need all three because any one of them can fail while the others hold.

What the checklist does not do is name the break. It is tempting — three conditions, three kinds of discontinuity, surely they line up — and it is the most common way to classify one wrong. They do not line up. Look at the three pictures below and count: two of them fail condition (2), and the first one can fail either (1) or (3) depending on whether the value is missing or merely misplaced. Three failure modes, three names, no matching.

The reason is worth a sentence, because it makes the right method obvious. The checklist is a test for continuity — it is built to return yes or no, and it stops as soon as it has the answer. A classification is a different question asked afterward, and it has a different source of evidence: the checklist settles whether the function broke; what the limit was doing settles how.

So run it in two stages. Stage one is the three conditions. If any fails, stage two is one more look — at the limit, not at the checklist:

Two habits fall straight out of this. First, never name a discontinuity before you have computed the one-sided limits; the name is that computation. Second, be suspicious of any rule that lets you skip it. “$f\left(a\right)$ is undefined, so it’s a hole” is the classic — and $\frac{1}{x}$ at $x=0$ is the counterexample every AP class meets in the first week. Condition (1) fails there, exactly as it does at a hole. It is a vertical asymptote, because the limit is what tells you, and the limit ran away.

The name “removable” is literal, and it is worth knowing why. In the first picture you could define $f\left(a\right)$ to be the limit value, plug the hole, and the function would be continuous — one point repaired and the problem is gone. Nothing you can do to a single point repairs any of the others, which is what the last column of the table was recording. The fourth kind, oscillating, is the one you cannot draw here: near $x=0$, $\sin (1∕x)$ crosses every value between $-1$ and $1$ infinitely often, so there is no picture with finitely many strokes in it.

INTERMEDIATE VALUE THEOREM

If $f$ is continuous on $[a,b]$ and $N$ is strictly between $f\left(a\right)$ and $f\left(b\right)$ —

— then $f\left(c\right)=N$ for some $c$ in the open interval $(a,b)$.

Those two strict inequalities are the price of the open interval, and they are cheap. If you let $N$ equal an endpoint value, the guarantee still holds but only on the closed interval: $f\left(x\right)=x$ on $\left[\mathrm{0,1}\right]$ with $N=0$ has exactly one solution, $c=0$, and it sits at the edge rather than inside. Write the endpoint values with $N$ caught strictly between them — as Problem 8 does, $-3<0<7$ — and the conclusion you want is the one you get.

Both hypotheses are load-bearing. Continuity is what forbids the function from skipping a value; without it the theorem says nothing at all, because a graph that is allowed to jump can step straight over $N$ (Problem 8 draws exactly that). And $N$ must actually be caught between the endpoint values — if it isn’t, the theorem has no opinion.

What it never does. The IVT is a one-way guarantee. When its hypotheses hold, the value must exist; when they don’t, it goes silent. Silence is not a “no.” A student who concludes “the IVT doesn’t apply, so there’s no root” has read a theorem backwards, and that specific error is worth its own exam question — which is Problem 8(b).

What it never tells you either: where $c$ is, or how many there are. It guarantees at least one. Finding it is a different problem with different tools.

The sentence that scores. A full-credit IVT justification names the continuity with a reason, shows the two endpoint values, and draws the conclusion:

Graders accept the argument without the theorem’s name attached — but citing it costs you four words and removes all doubt, so cite it.