Limits and Continuity Flashcards
10 cards from real AP CALC AB practice questions. Tap to flip, then mark Knew It or Still Learning — missed cards come back until you master them.
Read the first 10 Limits and Continuity flashcards as text
What is lim(x→2) (x² − 4)/(x − 2)?
Answer: 4
Factor the numerator as (x − 2)(x + 2) and cancel the common factor. That leaves lim(x→2) (x + 2) = 4.
What is lim(x→∞) (3x² + 5x)/(2x² − 1)?
Answer: 3/2
The numerator and denominator have the same degree, so the limit is the ratio of the leading coefficients, 3/2.
Let f(x) = kx + 1 for x ≤ 3 and f(x) = x² − 5 for x > 3. For what value of k is f continuous at x = 3?
Answer: 1
The right-hand limit at 3 is 3² − 5 = 4. The left piece must give 3k + 1 = 4, so k = 1.
What is lim(x→0) sin(5x)/(2x)?
Answer: 5/2
Rewrite the expression as (5/2)·[sin(5x)/(5x)]. Since sin(u)/u → 1 as u → 0, the limit is 5/2.
The function f is continuous on the closed interval [1, 4], with f(1) = −3 and f(4) = 6. Which statement must be true?
Answer: f(c) = 0 for some c in (1, 4)
By the Intermediate Value Theorem, a continuous f takes every value between −3 and 6 on (1, 4). Since 0 lies between them, f(c) = 0 for some c in (1, 4).
What is lim(x→3⁻) 2/(x − 3)?
Answer: −∞
As x approaches 3 from the left, x − 3 is a small negative number. Dividing 2 by it gives a large negative value, so the function decreases without bound.
Let f(x) = (x² − x − 6)/((x − 3)(x + 1)). At which value of x does f have a removable discontinuity?
Answer: 3
The numerator factors as (x − 3)(x + 2), so the factor (x − 3) cancels and leaves a hole at x = 3. At x = −1 the denominator still vanishes, which gives a vertical asymptote.
What is lim(x→0) (√(x + 9) − 3)/x?
Answer: 1/6
Multiply the numerator and denominator by the conjugate √(x + 9) + 3. The expression becomes 1/(√(x + 9) + 3), which approaches 1/6.
Let f(x) = x² for x ≠ 1 and f(1) = 3. Which statement is true?
Answer: lim(x→1) f(x) = 1, but f is not continuous at x = 1
Limits ignore the value at the point, so lim(x→1) f(x) = lim(x→1) x² = 1. Because this differs from f(1) = 3, f is not continuous at x = 1.
What is lim(x→−∞) (2x + 1)/√(x² + 4)?
Answer: −2
For large negative x, √(x² + 4) behaves like |x| = −x. The ratio then approaches 2x/(−x) = −2.
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