Limits & Continuity: AP Calculus BC Practice Problems

A two-sided limit exists only when both one-sided limits exist and agree. Check each jump in the graph.

At x=−1: the graph approaches 2 from the left (open circle) but the function actually equals 1 there (filled dot, matching where the middle piece begins) — the one-sided limits are 2 and 1, which disagree. The two-sided limit does not exist.

At x=1: the graph approaches 1 from the left (open circle) but jumps to 2 from the right (filled dot). Again the one-sided limits disagree, so the two-sided limit does not exist.

At x=2, the graph is a smooth, unbroken line — the limit exists there. The correct answer is D.

From the graph, as x approaches 1 from the left, f(x) approaches 1 (shown by the open circle at (1,1)). This one-sided limit exists — Statement I is true.

As x approaches 1 from the right, f(x) approaches 2 (shown by the filled dot at (1,2), where the third piece begins). This one-sided limit also exists — Statement II is true.

Since the left-hand limit (1) and right-hand limit (2) exist but are not equal, the two-sided limit does not exist — Statement III is false.

The correct answer is C.

The graph has two flaws: a jump at x=0 (the curve approaches one value from the left but jumps to a higher value at x=0 itself), and a removable discontinuity at x=2 (an open circle on the line, with the actual function value plotted separately below it).

[−2,0] includes x=0 as its right endpoint, and the function's actual value there (3) doesn't match the left-hand limit (1) — this interval is not continuous.

[2,3] includes x=2 as its left endpoint, where the isolated point doesn't match the surrounding line — not continuous. [0,3] includes x=2 as an interior point, which is exactly where the flaw lives — also not continuous.

The open interval (0,2) avoids both trouble spots entirely, following a single smooth line the whole way. The correct answer is B.

A jump discontinuity requires both one-sided limits to exist but disagree. At x=0, the left-hand limit is 1 and the function's value (which also equals the right-hand limit, since the right piece is defined right up to x=0) is 3 — both one-sided limits exist and clearly differ. This is a jump.

At x=2, the two-sided limit actually exists (both one-sided limits equal 5, since the line is unbroken there) — it's only the isolated function value (2) that's wrong. That makes x=2 a removable discontinuity, not a jump.

x=−2 and x=3 are just the domain's endpoints, with no discontinuity at either.

The correct answer is B.

As x→0⁺, 1/x→+∞, so arctan(1/x) → π/2. The expression under the radical approaches 5+2(π/2) = 5+π, so the right-hand limit is √(5+π).

As x→0⁻, 1/x→−∞, so arctan(1/x) → −π/2. The expression under the radical approaches 5−π (still positive, since π≈3.14<5), so the left-hand limit is √(5−π).

Since √(5+π) ≠ √(5−π), the left-hand and right-hand limits disagree.

The correct answer is D, the limit does not exist.

Since the left-hand and right-hand limits at x=2 both equal 4, the two-sided limit exists and equals 4 — Statement I is true.

The problem only gives information about f's behavior near x=2; nothing is stated about f anywhere else on its domain, so there isn't enough information to conclude f is continuous everywhere except x=2 — Statement II cannot be verified and is not necessarily true.

Since the limit exists (4) but f(2) is undefined, this is exactly the definition of a removable discontinuity — defining f(2)=4 would make f continuous there. Statement III is true.

The correct answer is C.

As x→0⁺, 1/x→+∞, so 5^1/x → ∞. The denominator grows without bound, so y → 0.

As x→0⁻, 1/x→−∞, so 5^1/x → 0. The denominator approaches 3+0=3, so y → 1/3.

Since the right-hand limit (0) and left-hand limit (1/3) are different, the two-sided limit does not exist.

The correct answer is D.

Vertical asymptotes occur where the denominator is zero but the numerator is not: x²−4=0 → x=±2. Since the numerator (6) is never zero, both are genuine vertical asymptotes.

For the horizontal asymptote, compare degrees: the numerator has degree 0, and the denominator has degree 2. Since the denominator's degree is larger, the horizontal asymptote is y=0 — the x-axis.

Both features exist simultaneously: the graph has vertical asymptotes at x=±2 and a horizontal asymptote along the x-axis.

The correct answer is C.

Factor the denominator to find candidate vertical asymptotes: 6x²−6x = 6x(x−1) = 0 → x=0, x=1. Check the numerator at each: at x=0 it's 4, and at x=1 it's 2+1+4=7 — neither is zero, so both are genuine vertical asymptotes.

Since the numerator and denominator have the same degree (both degree 2), the horizontal asymptote is the ratio of the leading coefficients: 2/6 = 1/3.

All three features hold at once: horizontal asymptote y=1/3, vertical asymptotes at x=0 and x=1.

The correct answer is C.

f(0) is explicitly defined to equal 4, so f(0) exists — Statement I is true.

For x≠0, simplify: (x²+3x)/x = x+3 (valid since x≠0 lets us cancel the common factor). So lim_x→0f(x) = lim_x→0(x+3) = 3. The limit exists — Statement II is true.

Continuity at x=0 requires f(0) = lim_x→0f(x). Here f(0)=4 but the limit is 3 — they don't match, so f is not continuous at x=0. Statement III is false (this is a removable discontinuity, not a point of continuity).

The correct answer is C.