Unit 1: Limits & Continuity | AP® Calculus AB Review

1. What a limit is

The limit $\lim_{x \to a} f(x) = L$ says: as $x$ gets arbitrarily close to $a$ (from either side, without touching $a$), the outputs $f(x)$ get arbitrarily close to $L$.

The single most important idea in this unit: the limit does not care what happens at $x = a$. The function can be undefined there, or defined with the "wrong" value, and the limit can still exist. The limit describes the approach, not the arrival.

Worked example

The graph of $f(x) = \dfrac{x^2 - 1}{x - 1}$ has a hole at $x = 1$. What is $\lim_{x \to 1} f(x)$?

Factor: $\dfrac{(x-1)(x+1)}{x-1} = x + 1$ for every $x \neq 1$.
Near $x = 1$, the function behaves exactly like the line $y = x + 1$.
So the limit is $1 + 1 = 2$, even though $f(1)$ itself is undefined.

Common mistake: answering "does not exist" whenever the function is undefined at the point. Holes are exactly where limits earn their keep.

2. Evaluating limits algebraically

Your toolbox, in the order you should reach for it:

Substitute first. If the function is continuous at $a$ (all polynomials, and rationals, roots, trig, exponentials wherever they are defined), the limit is just $f(a)$.

If substitution gives $\frac{0}{0}$, transform. That form means "hidden common factor." Three standard moves: factor and cancel, multiply by a conjugate (for square roots), or clear complex fractions with a common denominator.

Worked example (conjugate)

Evaluate $\lim_{x \to 9} \dfrac{\sqrt{x} - 3}{x - 9}$.

Substitution gives $\frac{0}{0}$. Multiply top and bottom by $\sqrt{x} + 3$.
Numerator becomes $x - 9$, which cancels the denominator.
Left with $\dfrac{1}{\sqrt{x} + 3}$. Substitute: $\dfrac{1}{6}$.

Common mistake: concluding $\frac{0}{0}$ means the limit is $0$, or that it does not exist. It means neither. It means "do more algebra."

3. The special trig limits

Two limits you must know cold:

$\lim_{x \to 0} \dfrac{\sin x}{x} = 1 \qquad\qquad \lim_{x \to 0} \dfrac{1 - \cos x}{x} = 0$

The workhorse move: force the argument of sine to match the denominator. $\dfrac{\sin(kx)}{x} = k \cdot \dfrac{\sin(kx)}{kx} \to k$.

Worked example

Evaluate $\lim_{x \to 0} \dfrac{\sin(3x)}{\sin(7x)}$.

Rewrite: $\dfrac{\sin(3x)}{3x} \cdot \dfrac{7x}{\sin(7x)} \cdot \dfrac{3}{7}$.
The first two factors each approach $1$.
The limit is $\dfrac{3}{7}$.

Common mistake: forgetting the coefficient. $\lim_{x \to 0} \frac{\sin(5x)}{x}$ is $5$, not $1$.

4. One-sided limits and when limits exist

$\lim_{x \to a^-}$ approaches from the left ($x < a$); $\lim_{x \to a^+}$ approaches from the right ($x > a$). The two-sided limit exists exactly when both one-sided limits exist and agree.

One-sided limits earn their keep at piecewise seams and absolute values. For absolute values, resolve the sign first: near $0$ from the left, $|x| = -x$.

Worked example

For $f(x) = \begin{cases} 2x + 1 & x < 3 \\ x^2 & x \geq 3 \end{cases}$, does $\lim_{x \to 3} f(x)$ exist?

Left: $\lim_{x \to 3^-} (2x + 1) = 7$.
Right: $\lim_{x \to 3^+} x^2 = 9$.
$7 \neq 9$, so the two-sided limit does not exist. That mismatch is a jump discontinuity.

Common mistake: using the piece that includes the equals sign for both directions. The equals sign tells you $f(3)$; the inequality tells you which piece governs each approach.

5. Limits at infinity and asymptotes

For rational functions as $x \to \pm\infty$, compare degrees: bottom bigger means limit $0$; equal degrees means ratio of leading coefficients; top bigger means $\pm\infty$. These limits are the horizontal asymptotes.

Vertical asymptotes come from denominator zeros that survive cancellation. A zero that cancels is a hole, not an asymptote. Always factor before classifying.

One trap deserves its own warning: for $x \to -\infty$, $\sqrt{x^2} = -x$ (positive!), which flips signs in square-root problems.

Worked example

Find all asymptotes of $f(x) = \dfrac{x - 3}{x^2 - 9}$.

Factor: $\dfrac{x - 3}{(x-3)(x+3)} = \dfrac{1}{x+3}$ for $x \neq 3$.
$x = 3$ canceled: that's a hole. $x = -3$ survives: vertical asymptote.
Degree of bottom exceeds top, so $y = 0$ is the horizontal asymptote.

Common mistake: declaring a vertical asymptote at every zero of the denominator without factoring first.

6. Continuity

$f$ is continuous at $a$ when three things hold: $f(a)$ is defined, $\lim_{x \to a} f(x)$ exists, and the two are equal. Break the checklist in different ways and you get the discontinuity types:

Removable (limit exists, wrong or missing value: a hole), jump (one-sided limits disagree), infinite (unbounded: a vertical asymptote).

The Intermediate Value Theorem: a function continuous on $[a, b]$ takes every value between $f(a)$ and $f(b)$. This is the exam's favorite way to ask "must a solution exist?"

Worked example

Find $k$ so that $f(x) = \begin{cases} kx^2 & x \leq 2 \\ 3x + 2 & x > 2 \end{cases}$ is continuous everywhere.

Each piece is a polynomial, continuous on its own domain. Only the seam $x = 2$ can fail.
Match the pieces there: $4k = 8$.
$k = 2$ makes both sides meet at $8$, so $f$ is continuous.

Common mistake: on IVT questions, forgetting to state that the function is continuous. On the FRQ, the word "continuous" is usually worth the point.

7. The squeeze theorem

If $g(x) \leq f(x) \leq h(x)$ near $a$ and both $g$ and $h$ approach the same value $L$, then $f$ approaches $L$ too. Use it when a bounded, oscillating factor (like $\sin\frac{1}{x}$) is multiplied by something shrinking to $0$.

Worked example

Evaluate $\lim_{x \to 0} x^2 \sin\left(\tfrac{1}{x}\right)$.

$-1 \leq \sin(1/x) \leq 1$ for all $x \neq 0$.
Multiply by $x^2$: $-x^2 \leq x^2 \sin(1/x) \leq x^2$.
Both bounds go to $0$, so the limit is $0$.

Common mistake: assuming any oscillation kills a limit. Oscillation with shrinking amplitude is exactly where the squeeze theorem shines.

Unit 1 practice set: 45 questions

Every question exam-style, every solution step-by-step. You'll get a topic report at the end showing exactly what to review.