Limits and Continuity Flashcards

What is lim(x→3) (x² − 9)/(x − 3)?

Answer: 6

Factor the numerator as (x − 3)(x + 3) and cancel (x − 3). The limit is then lim(x→3) (x + 3) = 6.

What is lim(x→0) sin(5x)/(2x)?

Answer: 5/2

Rewrite as (5/2)·[sin(5x)/(5x)]. Since sin(u)/u → 1 as u → 0, the limit is 5/2.

Let f(x) = kx + 2 for x ≤ 2 and f(x) = x² − k for x > 2. For what value of k is f continuous at x = 2?

Answer: 2/3

Continuity requires the left and right limits to equal f(2). So 2k + 2 = 4 − k, which gives k = 2/3.

What is lim(x→∞) (3x² − 5x + 1)/(2x² + 7)?

Answer: 3/2

The numerator and denominator have the same degree, so the limit is the ratio of the leading coefficients, 3/2.

What is lim(x→2⁻) 1/(x − 2)?

Answer: −∞

For x slightly less than 2, x − 2 is a small negative number, so 1/(x − 2) is a large negative number. The limit is −∞.

A function f is continuous on [1, 4] with f(1) = −3 and f(4) = 5. Which statement must be true?

Answer: f(c) = 0 for at least one c in (1, 4)

0 lies between f(1) = −3 and f(4) = 5, so the Intermediate Value Theorem guarantees at least one c in (1, 4) with f(c) = 0.

What type of discontinuity does f(x) = (x² − x − 6)/(x − 3) have at x = 3?

Answer: Removable discontinuity

The numerator factors as (x − 3)(x + 2), so f(x) = x + 2 for x ≠ 3. The limit is 5 even though f(3) is undefined, which makes the discontinuity removable.

What is lim(x→0) (1 − cos x)/x²?

Answer: 1/2

Multiply by (1 + cos x)/(1 + cos x) to get sin²x / [x²(1 + cos x)]. This approaches 1·(1/2) = 1/2.

What is lim(x→2) |x − 2|/(x − 2)?

Answer: Does not exist

From the right the expression equals 1, and from the left it equals −1. The one-sided limits differ, so the two-sided limit does not exist.

What are the horizontal asymptotes of f(x) = 2x/√(x² + 1)?

Answer: y = 2 and y = −2

As x → ∞, √(x² + 1) ≈ x, so f → 2. As x → −∞, √(x² + 1) ≈ |x| = −x, so f → −2.