Limits and Continuity — High School Calculus practice questions

Evaluate lim_{x→4} (√x − 2)/(x − 4).

1. 1/4
2. 0
3. 1/2
4. Does not exist

Answer: A — 1/4

A) Correct — Multiply by the conjugate: (x − 4)/[(x − 4)(√x + 2)] = 1/(√x + 2) → 1/4. B) Stops at the 0/0 form. 0/0 is indeterminate: it signals that more algebra is needed, not that the answer is 0. C) Forgets the + 2 in the conjugate and uses 1/√x. D) Treats 0/0 as a final answer. The form is indeterminate, and this limit does exist.

Evaluate lim_{x→∞} e^(−x).

1. 1
2. −∞
3. e
4. 0

Answer: D — 0

D) Correct — e^(−x) = 1/e^x, and e^x grows without bound. A) Evaluates e^0 instead of letting x grow. B) Treats the negative exponent as a negative value; e^(−x) is always positive. C) Ignores the exponent.

Evaluate lim_{x→2} (x² − 4)/(x − 2).

1. 0
2. 4
3. Does not exist
4. 2

Answer: B — 4

B) Correct — Factor: (x − 2)(x + 2)/(x − 2) = x + 2 for x ≠ 2, so the limit is 2 + 2 = 4. A) Stops at the 0/0 form. 0/0 is indeterminate: it signals that more algebra is needed, not that the answer is 0. C) Treats 0/0 as a final answer. The form is indeterminate, and this limit does exist. D) Cancels incorrectly and is left with x instead of x + 2.

Evaluate lim_{x→0} x² sin(1/x).

1. Does not exist
2. 1
3. 0
4. ∞

Answer: C — 0

C) Correct — Since −x² ≤ x² sin(1/x) ≤ x² and both bounds go to 0, the squeeze theorem gives 0. A) sin(1/x) oscillates, but the x² factor squeezes the product to 0. B) Confuses this with lim (sin x)/x = 1. D) sin(1/x) stays between −1 and 1, so nothing blows up.

For f(x) = |x|/x, what is lim_{x→0⁻} f(x)?

1. 1
2. 0
3. −1
4. Does not exist

Answer: C — −1

C) Correct — For x < 0, |x| = −x, so f(x) = −x/x = −1. A) Is the right-hand limit, for x > 0. B) Substitutes x = 0 into |x| and ignores the 0 in the denominator. D) The two-sided limit does not exist, but this one-sided limit does.

Evaluate lim_{x→∞} (5x + 1)/(x² + 4).

1. 5
2. ∞
3. 1/4
4. 0

Answer: D — 0

D) Correct — The denominator has the higher degree, so the fraction shrinks to 0. A) Divides the leading coefficients even though the degrees are different. B) Reverses the degree rule; ∞ happens when the numerator's degree is higher. C) Uses the constant terms, which become negligible as x → ∞.

Where does f(x) = (x + 2)/(x² − 4) have a vertical asymptote?

1. x = 2 and x = −2
2. x = −2 only
3. y = 0
4. x = 2 only

Answer: D — x = 2 only

D) Correct — x² − 4 = (x − 2)(x + 2). The (x + 2) cancels, leaving a hole at x = −2 and an asymptote only at x = 2. A) Misses that x + 2 cancels, which makes x = −2 a hole. B) Reverses which factor cancels. C) Is the horizontal asymptote, not a vertical one.

A function f is continuous at x = a when:

1. f(a) is defined
2. lim_{x→a} f(x) exists
3. f(a) is defined, lim_{x→a} f(x) exists, and the two are equal
4. The graph has no sharp corner at a

Answer: C — f(a) is defined, lim_{x→a} f(x) exists, and the two are equal

C) Correct — All three conditions are needed. A) Is only one of the three conditions. B) Is only one of the three conditions. D) Corners are allowed: |x| is continuous at 0. Corners affect differentiability.

Evaluate lim_{x→0} (sin x)/x.

1. 0
2. Does not exist
3. ∞
4. 1

Answer: D — 1

D) Correct — This is a standard limit: near 0, sin x ≈ x, so the ratio approaches 1. A) Notices sin 0 = 0 and assumes the whole fraction is 0, but the denominator is also 0. B) Treats 0/0 as a final answer. The form is indeterminate, and this limit does exist. C) Treats the 0 in the denominator as division by zero; the numerator shrinks just as fast.

State the Intermediate Value Theorem.

If f is continuous on [a, b] and N is between f(a) and f(b), then there exists c in [a, b] with f(c) = N.

Evaluate lim_{h→0} [(2 + h)² − 4]/h.

Expand: (4 + 4h + h² − 4)/h = (4h + h²)/h = 4 + h → 4.

Evaluate lim_{x→∞} (3x² + 2x)/(x² − 5).

3 (ratio of leading coefficients, since degrees are equal).

What does it mean for f to be continuous at a?

f(a) defined, lim_{x→a} f(x) exists, and lim = f(a).