Test yourself on Limits with 12 original AP Exams practice questions. Pick an answer to see instant feedback and a full explanation.
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Answer the questions below — you get instant feedback and a full explanation for each.
1. Evaluate lim(x→3) (x² - 9)/(x - 3).
Explanation. Factor: (x²-9)/(x-3) = (x-3)(x+3)/(x-3) = x+3 for x≠3. As x→3, this approaches 3+3 = 6.
2. Evaluate lim(x→0) (sin 5x)/x.
Explanation. Rewrite as 5·(sin5x)/(5x). Since lim(u→0) (sin u)/u = 1 with u=5x, the limit is 5·1 = 5.
3. Evaluate lim(x→∞) (3x² + 2x - 1)/(5x² - 4).
Explanation. For a rational function where numerator and denominator have equal degree, the limit equals the ratio of leading coefficients: 3/5.
4. Evaluate lim(x→0) (√(x+4) - 2)/x.
Explanation. Multiply by the conjugate: [(√(x+4)-2)(√(x+4)+2)]/[x(√(x+4)+2)] = (x+4-4)/[x(√(x+4)+2)] = 1/(√(x+4)+2). As x→0 this is 1/(2+2) = 1/4.
5. For f(x) defined piecewise as f(x)=x+1 for x<2 and f(x)=3x-3 for x≥2, what is lim(x→2) f(x)?
Explanation. Left limit: 2+1 = 3. Right limit: 3(2)-3 = 3. Both sides equal 3, so the limit exists and equals 3.
6. Evaluate lim(x→2⁺) 1/(x - 2).
Explanation. As x approaches 2 from the right, x-2 is a small positive number, so 1/(x-2) grows without bound. The limit is +∞ (an infinite limit indicating a vertical asymptote).
7. Which statement correctly describes when lim(x→a) f(x) exists?
Explanation. A two-sided limit exists if and only if the left-hand and right-hand limits both exist and are equal. The limit can exist even if f(a) is undefined or f is not continuous/differentiable there.
8. Evaluate lim(x→∞) (2x + 1)/(√(x² + 3)).
Explanation. Divide top and bottom by x (for x>0, √(x²+3)=x√(1+3/x²)): (2 + 1/x)/√(1+3/x²) → (2+0)/√1 = 2.
9. Use the Squeeze Theorem: what is lim(x→0) x²·cos(1/x)?
Explanation. Since -1 ≤ cos(1/x) ≤ 1, we have -x² ≤ x²cos(1/x) ≤ x². Both -x² and x² approach 0 as x→0, so by the Squeeze Theorem the limit is 0.
10. Evaluate lim(x→0) (1 - cos x)/x².
Explanation. A standard limit: (1-cos x)/x² → 1/2. This follows from multiplying by (1+cos x)/(1+cos x) giving sin²x/[x²(1+cos x)] → (1)(1/(1+1)) = 1/2.
11. What type of discontinuity does f(x) = (x²-1)/(x-1) have at x=1?
Explanation. f(x) simplifies to x+1 for x≠1, so lim(x→1) f(x) = 2 exists, but f(1) is undefined. A finite limit existing where the function is undefined indicates a removable (hole) discontinuity.
12. Evaluate lim(x→0) (e^(2x) - 1)/x.
Explanation. Rewrite as 2·(e^(2x)-1)/(2x). Since lim(u→0)(e^u-1)/u = 1 with u=2x, the limit is 2·1 = 2. (Equivalently, this is the derivative of e^(2x) at 0.)
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FAQ
When should I use algebraic manipulation versus L'Hôpital's rule on limits?
On AP Calculus AB, factoring, conjugate multiplication, and known special limits (like sin x/x) are the expected first methods. L'Hôpital's rule applies only to 0/0 or ∞/∞ indeterminate forms and is part of the curriculum, but make sure you verify the indeterminate form first before applying it.
How do I handle limits at infinity for rational functions?
Compare the degrees of numerator and denominator: if numerator degree < denominator degree, the limit is 0; if equal, it's the ratio of leading coefficients; if numerator degree > denominator degree, the limit is ±∞ (no horizontal asymptote).
What's the difference between a limit being infinite and not existing?
A limit of +∞ or -∞ describes specific unbounded behavior (often a vertical asymptote) and is acceptable AP language. A limit 'does not exist' (DNE) typically means the left and right sides disagree or the function oscillates without settling. Read questions carefully—they often want '+∞' rather than just 'DNE.'