AP Calculus AB Practice: Limits & Continuity (17 Step-by-Step Explanations)

AP Calculus AB · Limits & Continuity

Limits & Continuity: AP Calculus AB Practice Problems

Limits are the foundation everything else in AP Calculus AB is built on, and the exam tests them from nearly every angle: algebraic limits that require factoring or rationalizing to resolve a 0/0 form, limits at infinity that depend on comparing the degrees of a rational function's numerator and denominator, limits of piecewise and exponential functions, and limits read directly off a graph. Continuity questions build on the same ideas, asking you to classify a discontinuity as removable, a jump, or an infinite (vertical asymptote) discontinuity based on how the one-sided limits behave.

A recurring theme across these problems is that direct substitution is only the first step, not the whole solution. When substitution produces an indeterminate 0/0 form, the path forward depends on the structure of the expression: polynomials usually factor and cancel a common term, expressions with square roots usually need to be multiplied by a conjugate, and expressions built from known trig limits (like sin x/x) can often be rewritten in terms of those standard results. For limits at infinity, the End Behavior of a rational function is governed entirely by its highest-degree terms — comparing the degree of the numerator to the degree of the denominator tells you immediately whether the limit is 0, a finite ratio of leading coefficients, or infinite.

Below are 17 AP Calculus AB-style practice problems covering limits at infinity, algebraic limit techniques, the Squeeze Theorem, the Intermediate Value Theorem, continuity and discontinuity classification, and reading limits from graphs. Work through each problem, select your answer, and check whether you got it right — every problem includes a complete, step-by-step explanation. Once you've worked through these, keep building your skills with thousands more official-style questions in our free AP Calculus AB QBank.

Intermediate Value Theorem

Find the interval for which the Intermediate Value Theorem guarantees the existence of at least one number c in that interval for which f(c) = 0, if f(x) = −x² − 5x − 2.

Full Explanation

The IVT guarantees a root inside an interval only when f changes sign between the two endpoints. Evaluate f at each candidate endpoint:

f(−8) = −64+40−2 = −26, f(−6) = −36+30−2 = −8, f(−4) = −16+20−2 = 2, f(−3) = −9+15−2 = 4, f(−1) = −1+5−2 = 2.

Check each interval for a sign change: [−8,−6] goes −26 to −8 (no change); [−6,−4] goes −8 to 2 (sign change!); [−4,−3] goes 2 to 4 (no change); [−3,−1] goes 4 to 2 (no change).

Only [−6,−4] has f changing from negative to positive, so the IVT guarantees a root there. The correct answer is B.

Intermediate Value Theorem

Find the interval for which the Intermediate Value Theorem guarantees the existence of at least one number c in that interval for which f(c) = 0, if f(x) = 3x² − x − 10.

Full Explanation

Evaluate f at each candidate endpoint: f(−1) = 3+1−10 = −6, f(0) = −10, f(1) = 3−1−10 = −8, f(3) = 27−3−10 = 14, f(6) = 108−6−10 = 92.

Check each interval: [−1,0] goes −6 to −10 (no change); [0,1] goes −10 to −8 (no change); [1,3] goes −8 to 14 (sign change!); [3,6] goes 14 to 92 (no change).

Only [1,3] has f changing from negative to positive. The correct answer is C.

Limit at infinity · exponential functions

lim_x→−∞ (−4+7^x)/(5−8^x)

Full Explanation

For any base b > 1, b^x → 0 as x → −∞. This applies to both 7^x and 8^x here.

As x → −∞, the expression approaches (−4+0)/(5−0) = −4/5.

The correct answer is C.

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Limit at infinity · radical over polynomial

lim_x→−∞ √(9x⁶+3x²) / (7x³+4x)

Full Explanation

Factor x⁶ out from under the radical: √(9x⁶+3x²) = √(x⁶(9+3/x⁴)) = |x³|√(9+3/x⁴). Since x → −∞ means x is negative, |x³| = −x³.

Divide numerator and denominator by x³: [−x³√(9+3/x⁴)] / x³ = −√(9+3/x⁴) for the numerator, and (7x³+4x)/x³ = 7+4/x² for the denominator.

As x → −∞, both fraction terms vanish: numerator → −√9 = −3, denominator → 7.

The limit is −3/7. The correct answer is A. The negative sign is easy to miss — it comes directly from x being negative as it approaches −∞.

Limit at infinity · exponential functions

lim_x→−∞ (6−4^x)/(3+5^x)

Full Explanation

As x → −∞, both 4^x and 5^x approach 0, since any exponential with base greater than 1 shrinks toward 0 as x becomes very negative.

The expression approaches (6−0)/(3+0) = 6/3 = 2.

The correct answer is D.

Limit at infinity · comparing polynomial degrees

lim_x→∞ (7−9x³)/(x⁴−1)

Full Explanation

When the degree of the denominator is greater than the degree of the numerator, the denominator grows faster and the fraction shrinks toward 0.

Here the numerator has degree 3 and the denominator has degree 4, so the denominator dominates as x → ∞.

The correct answer is C, 0.

Limit · factor and cancel

lim_t→−3 (t²+11t+24)/(t²−9)

Full Explanation

Direct substitution gives 0/0, so factor both the numerator and denominator. The numerator factors as (t+8)(t+3) (8 × 3 = 24, 8+3=11), and the denominator factors as (t−3)(t+3).

Cancel the common factor (t+3), valid since t ≠ −3 near the limit: (t+8)/(t−3).

Substitute t = −3: (−3+8)/(−3−3) = 5/(−6) = −5/6.

The correct answer is B.

7 down, 10 to go — keep practicing.

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Limit · rewriting in terms of a known trig limit

lim_x→0 tan x / (x²−4x)

Full Explanation

Factor the denominator and split off the well-known limit tan x / x → 1 as x → 0:

tan x / (x²−4x) = tan x / [x(x−4)] = (tan x / x) · (1/(x−4))

As x → 0: tan x / x → 1, and 1/(x−4) → 1/(0−4) = −1/4.

Multiplying these limits together: 1 × (−1/4) = −1/4. The correct answer is D.

Limit · recognizing a difference-of-squares-style cancellation

lim_x→10 (10−x)² / (x−10)

Full Explanation

Notice that (10−x) = −(x−10), so (10−x)² = (x−10)² (squaring removes the sign difference).

The expression becomes (x−10)² / (x−10) = (x−10), valid for x ≠ 10.

As x → 10, (x−10) → 0.

The correct answer is D.

Limit · rationalizing a radical expression

lim_x→5 [√(4x+5) − √(5x)] / (x−5)

Full Explanation

Direct substitution gives 0/0 (both radicals equal 5 at x=5), so multiply by the conjugate √(4x+5)+√(5x) over itself:

The numerator becomes a difference of squares: (4x+5)−5x = 5−x = −(x−5).

The expression is now −(x−5) / [(x−5)(√(4x+5)+√(5x))] = −1/(√(4x+5)+√(5x)) for x ≠ 5.

As x → 5: −1/(√25+√25) = −1/(5+5) = −1/10. The correct answer is C.

Limit · rewriting trig functions in terms of sine and cosine

lim_x→0 tan x · csc x

Full Explanation

Rewrite both trig functions in terms of sine and cosine: tan x = sin x / cos x and csc x = 1/sin x.

Multiply them together: (sin x / cos x) · (1/sin x) = 1/cos x, valid wherever sin x ≠ 0.

As x → 0: 1/cos(0) = 1/1 = 1.

The correct answer is C.

11 down, 6 to go.

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Limit at negative infinity · comparing polynomial degrees

lim_x→−∞ (−3x³+4x−9)/(−5x²+6x−1)

Full Explanation

The numerator has degree 3 and the denominator has degree 2 — since the numerator's degree is higher, the limit is infinite, not a finite ratio.

Focus on the leading terms: (−3x³)/(−5x²) = (3/5)x.

As x → −∞, (3/5)x → −∞ (a positive coefficient times x, as x becomes very negative).

The correct answer is A.

Classifying a discontinuity

Let f be the function given by f(x) = (x−5) / (7|x−5|). Which of the following is true?

Full Explanation

Split the absolute value by cases. For x > 5: |x−5| = x−5, so f(x) = (x−5)/(7(x−5)) = 1/7 (constant).

For x < 5: |x−5| = −(x−5), so f(x) = (x−5)/(−7(x−5)) = −1/7 (constant).

The right-hand limit is 1/7 and the left-hand limit is −1/7. Since these are two different finite values, this is a jump discontinuity — not removable (the one-sided limits don't even agree) and not a vertical asymptote (both sides are finite, not infinite).

The correct answer is B.

One-sided limit of a piecewise function

lim_x→4⁻ f(x) if f(x) = 6x+1 for x < 4, and f(x) = 4/(x+7) for x ≥ 4

Full Explanation

The notation x → 4⁻ means x approaches 4 from the left (values less than 4), so use the piece of the function defined for x < 4: f(x) = 6x+1.

Substitute x = 4 into that piece: 6(4)+1 = 24+1 = 25.

The correct answer is B. (Note this is only the one-sided limit — it doesn't require checking the other piece of the function, since we're only approaching from the left.)

14 down, 3 to go.

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Limit · combining fractions before substituting

lim_x→0 [ 1/(x+7) − 1/7 ] / x

Full Explanation

Combine the two fractions in the numerator over a common denominator first: 1/(x+7) − 1/7 = [7−(x+7)] / [7(x+7)] = −x / [7(x+7)].

Now divide this whole expression by x (the outer denominator): [−x/(7(x+7))] / x = −1/[7(x+7)], valid for x ≠ 0.

As x → 0: −1/[7(0+7)] = −1/49.

The correct answer is A.

Finding where an error occurs in a Squeeze Theorem proof

A student attempts to solve lim_x→0 [x⁶sin(1/x)+2] using the squeeze theorem. The steps of the student's work are shown below. In which of the following steps does an error first appear?

Step 1: −2x⁶ ≤ x⁶sin(1/x)+2 ≤ 2x⁶
Step 2: lim_x→0 −2x⁶ = 0
Step 3: lim_x→0 2x⁶ = 0
Step 4: ∴ lim_x→0 [x⁶sin(1/x)+2] = 0

Full Explanation

Since sin(1/x) is always between −1 and 1, multiplying by x⁶ (which is always ≥ 0) gives −x⁶ ≤ x⁶sin(1/x) ≤ x⁶.

To bound the full expression x⁶sin(1/x)+2, the "+2" needs to be added to every part of that inequality: −x⁶+2 ≤ x⁶sin(1/x)+2 ≤ x⁶+2.

The student's Step 1 instead shows bounds of −2x⁶ and 2x⁶ — incorrectly doubling the x⁶ coefficient instead of adding the constant 2 to the bounds. This is a setup error, not an arithmetic slip in a later step.

The correct answer is A, Step 1. (With the correct bounds, both outer limits would approach 2 as x → 0, and the true answer would be 2, not 0.)

Limits of combined functions from two graphs

The graphs of the functions f and g are shown in the figures below. Which of the following statements is false?

Graph of f

Graph of g

Full Explanation

From the graph of f: the open circle at (2,4) means lim_x→2 f(x) = 4 (both sides approach 4, even though the filled dot shows the actual value f(2) = −1 is different). Statement A is true.

From the graph of g: the filled dot shows the line reaches g(−4) = −2 from the left, while the open circle shows the curve on the right approaches −4 (but doesn't reach it) as x → −4⁺. Since the left-hand limit (−2) and right-hand limit (−4) disagree, lim_x→−4 g(x) does not exist. Statement C is true.

For statement B: as x → 2, the inner input x−6 → −4, so g(x−6) approaches g's two different one-sided values (−2 from the left, −4 from the right) depending on the direction of approach, while f(x) → 4 from both sides. The product approaches −8 from one side and −16 from the other — these disagree, so the limit does not exist. Statement B is true.

For statement D: as x → −4, the inner input x+6 → 2, so f(x+6) → 4 consistently from both sides. But g(x) still approaches two different values (−2 from the left, −4 from the right). The product approaches (−2)(4) = −8 from the left and (−4)(4) = −16 from the right — these disagree, so this limit does not exist, contradicting the statement.

Statement D claims the limit exists, but it does not. The correct answer is D — this is the false statement.

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