AP Calculus AB Limits and Continuity — Worked Answer Explanations

Unit 1 · 12% of the AP exam · 8 questions explained

Below is a complete answer key for our AP Calculus AB Limits and Continuity practice questions. For each question you'll find the correct choice, a full written explanation of how to get there, and — for every wrong answer — a short note on exactly why it's tempting and where it goes wrong. Reading these straight through is one of the fastest ways to find the gaps in a unit before exam day.

Prefer to test yourself first? Take the timed Limits and Continuity practice test and come back here to review, or head back to the Limits and Continuity unit overview.

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Question 1 · Easy
Evaluate $x \to 3 lim \frac{x ^{2} - 9}{x - 3}$ .
- A
  $0$
  Why not A: Plugged in $x = 3$ into the unsimplified fraction.
- B
  $3$
  Why not B: Used $x$ as the answer rather than evaluating $x + 3$ .
- C
  $6$ Correct
- D
  Does not exist.
  Why not D: $0/0$ is indeterminate, not nonexistent.
Explanation
Factor: $\frac{x ^{2} - 9}{x - 3} = \frac{( x - 3 ) ( x + 3 )}{x - 3} = x + 3$ . Limit as $x \to 3$ is $6$ .

Key takeaway
Indeterminate $0/0$ forms — try factoring or other algebraic manipulation first.
Question 2 · Easy
Evaluate $x \to 0 lim \frac{sin ( 3 x )}{x}$ .
- A
  $0$
  Why not A: Plugged in $x = 0$ directly.
- B
  $1$
  Why not B: Used the basic identity without the coefficient.
- C
  $3$ Correct
- D
  $\infty$
  Why not D: The limit is finite.
Explanation
$\frac{sin ( 3 x )}{x} = 3 \cdot \frac{sin ( 3 x )}{3 x}$ . As $x \to 0$ , $\frac{sin ( 3 x )}{3 x} \to 1$ , so the limit is $3$ .

Key takeaway
$\lim_{u \to 0} \sin(u)/u = 1$ — pull out coefficients to apply this identity.
Question 3 · Easy
Find $x \to \infty lim \frac{2 x ^{2} - 5 x + 1}{3 x ^{2} + x - 7}$ .
- A
  $0$
  Why not A: Lower degree denominator answer.
- B
  $\frac{2}{3}$ Correct
- C
  $\frac{3}{2}$
  Why not C: Inverted the leading-coefficient ratio.
- D
  $\infty$
  Why not D: Higher degree numerator answer.
Explanation
Degrees of numerator and denominator are equal; limit is the ratio of leading coefficients: $2/3$ .

Key takeaway
For rational functions at infinity: equal degrees → ratio of leading coefficients.
Question 4 · Easy
If $x \to 4^{-} lim f (x) = 5$ and $x \to 4^{+} lim f (x) = 7$ , then:
- A
  $lim_{x \to 4} f (x) = 6$ .
  Why not A: Two-sided limit doesn't exist if one-sided limits differ.
- B
  $lim_{x \to 4} f (x) = 12$ .
  Why not B: Limits aren't summed.
- C
  $lim_{x \to 4} f (x)$ does not exist.Correct
- D
  $f$ is continuous at $4$ .
  Why not D: Discontinuous due to mismatched one-sided limits.
Explanation
The two-sided limit exists iff both one-sided limits exist AND are equal. Here $5 \neq = 7$ , so $lim_{x \to 4} f (x)$ does not exist.

Key takeaway
Two-sided limits require matching one-sided limits.
Question 5 · Medium
If $f (x) = \frac{x ^{2} - 4}{x - 2}$ for $x \neq = 2$ and $f (2) = 5$ , then $f$ is:
- A
  Continuous everywhere.
  Why not A: $f (2) \neq = lim_{x \to 2} f (x)$ .
- B
  Continuous everywhere except at $x = 2$ .Correct
- C
  Discontinuous everywhere.
  Why not C: It's continuous at every other $x$ .
- D
  Has a vertical asymptote at $x = 2$ .
  Why not D: $x = 2$ is a removable, not infinite, discontinuity.
Explanation
$lim_{x \to 2} f (x) = 4$ (after factoring), but $f (2) = 5$ . So $f$ has a jump (non-removable) discontinuity at $x = 2$ but is continuous elsewhere.

Key takeaway
A function is continuous at $a$ iff $\lim_{x \to a} f(x) = f(a)$ — both must match.
Question 6 · Medium
Evaluate $x \to 0 lim \frac{1 - cos x}{x ^{2}}$ .
- A
  $0$
  Why not A: Numerator and denominator both go to 0; not necessarily 0.
- B
  $\frac{1}{2}$ Correct
- C
  $1$
  Why not C: Off by a factor of 2.
- D
  Does not exist.
  Why not D: Limit exists.
Explanation
Standard limit. Multiply numerator and denominator by $1 + cos x$ : $\frac{( 1 - cos ^{2} x )}{x ^{2} ( 1 + cos x )} = \frac{sin ^{2} x}{x ^{2} ( 1 + cos x )} \to \frac{1}{2}$ as $x \to 0$ (since $sin x / x \to 1$ and $1 + cos x \to 2$ ).

Key takeaway
Standard limit: $\lim_{x\to 0}\dfrac{1-\cos x}{x^2} = \dfrac{1}{2}$.
Question 7 · Medium
Apply the squeeze theorem to evaluate $x \to 0 lim x^{2} sin (\frac{1}{x})$ .
- A
  $0$ Correct
- B
  $1$
  Why not B: Used $sin u / u$ identity which doesn't apply here.
- C
  $\infty$
  Why not C: Bounded sine times zero gives zero.
- D
  Does not exist.
  Why not D: Squeeze theorem confirms it does exist.
Explanation
$- x^{2} \leq x^{2} sin (1/ x) \leq x^{2}$ . As $x \to 0$ , both bounds go to 0, so the middle expression also goes to 0.

Key takeaway
Squeeze theorem: bounded function times zero-going function is zero.
Question 8 · Hard
The function $f (x) = {x^{2} a x + b x < 1 x \geq 1$ is continuous and differentiable at $x = 1$ . What are $a$ and $b$ ?
- A
  $a = 1$ , $b = 0$
  Why not A: Continuous but not differentiable.
- B
  $a = 2$ , $b = - 1$ Correct
- C
  $a = 2$ , $b = 1$
  Why not C: Slope match but $f (1)$ doesn't match.
- D
  $a = - 1$ , $b = 2$
  Why not D: Misapplied derivative matching.
Explanation
Continuity: $1 = a (1) + b$ , so $a + b = 1$ . Differentiability: $\frac{d}{d x} x^{2}_{x = 1} = 2 = a$ . Then $b = 1 - 2 = - 1$ .

Key takeaway
Match values for continuity, match derivatives for differentiability at the boundary.