Connecting Infinite Limits and Vertical Asymptotes: Practice Questions (Calculus AB 1.14)

Answer: (A). Substitution gives 6/0: nonzero over 0, so the expression is unbounded near 2. The top tends to 6, which is positive. For x slightly less than 2, x − 2 is a small negative number. Positive ÷ small negative = large negative. So the limit is −∞.

• (B) treats a nonzero number over 0 as 0. Dividing by something tiny makes the result huge, not 0.
• (C) “cancels” x from 3x and x − 2. The x in x − 2 is part of a sum, not a factor, so it cannot cancel.
• (D) is the right-hand limit. It ignores that x − 2 is negative when x < 2.

Answer: (C). At x = 1, (x + 2)/(x² − 1) gives 3/0. Nothing cancels, because x + 2 is not 0 at x = 1. The bottom factors as (x − 1)(x + 1), and for x slightly more than 1 it is a small positive number, so lim (x → 1⁺) f(x) = +∞. (From the left it is −∞.) One infinite one-sided limit is enough, so x = 1 is a vertical asymptote. This function also has a vertical asymptote at x = −1.

• (A) gives 0/0 at x = 1. The factor (x − 1) cancels to leave x + 1, so the limit is 2 and the graph has a hole at (1, 2), not an asymptote.
• (B) has a bottom, x² + 1, that is never 0. At x = 1 the function simply equals 0.
• (D) has ln 1 = 0, so the graph crosses the x-axis at x = 1. Its vertical asymptote is at x = 0, where lim (x → 0⁺) ln x = −∞.

Answer: (B). The top tends to 3 − 5 = −2, which is negative. The bottom (x − 3)² is a square, so it is a small positive number on both sides of 3. Negative ÷ small positive = large negative, on both sides. The one-sided limits agree, so lim (x → 3) f(x) = −∞.

• (A) gets the size right but ignores the negative top.
• (C) is the pattern for an odd power such as (x − 3). A squared factor does not change sign at 3.
• (D) assumes every zero of the bottom gives a hole. A hole needs 0/0 with a factor that cancels; here the top is −2, not 0.

Answer: (D). The top tends to 1. The bottom tends to ln 1 = 0. For x slightly more than 1, ln x is a small positive number. Positive ÷ small positive = large positive, so the limit is +∞. (This shows x = 1 is a vertical asymptote of y = x/ln x.)

• (A) uses the wrong sign for ln x. ln x < 0 only for 0 < x < 1; to the right of 1 it is positive.
• (B) treats 1/0 as 0.
• (C) uses ln 1 = 1. In fact ln 1 = 0, because e⁰ = 1.

(a) From the left the values are negative and grow in size by about 10 times each step closer to 5, so they appear unbounded: lim (x → 5⁻) g(x) = −∞. From the right they are positive and grow the same way: lim (x → 5⁺) g(x) = +∞.

(b) The line x = 5 appears to be a vertical asymptote, because at least one one-sided limit at 5 (in fact both) appears infinite.

(c) The two sides behave differently: −∞ from the left, +∞ from the right. A two-sided statement “= ∞” would claim the values are large and positive on both sides. The correct description is two separate one-sided limits.

(d) At x = 5 the top is 5 − 2 = 3, which is positive, and the bottom is 0: nonzero over 0. For x > 5, x − 5 > 0, so g(x) → +∞. For x < 5, x − 5 < 0, so g(x) → −∞. This matches (a). The table only suggested the behaviour; the sign check confirms it.

Suggested mark points (3): 1 for both one-sided limits with correct signs; 1 for naming a vertical asymptote with a limit as the reason; 1 for (c) and (d) together: the sides disagree, and the sign check of 3/(small positive or negative) is shown.

(a) The bottom is x² − 4 = (x − 2)(x + 2), so f is undefined at x = 2 and x = −2. The top is 2x² − 2x − 4 = 2(x² − x − 2) = 2(x − 2)(x + 1). Cancel (x − 2), valid for x ≠ 2:

f(x) = 2(x + 1)/(x + 2) for x ≠ 2.

• At x = 2: the simplified form gives 2(3)/4 = 3/2, a finite limit. So there is a hole at (2, 3/2).
• At x = −2: the simplified form gives 2(−1)/0 = −2/0, nonzero over 0. So there is a vertical asymptote at x = −2.

(b) Near x = −2 the top 2(x + 1) is close to −2, which is negative.

• x → −2⁺: x + 2 is small and positive. Negative ÷ small positive → lim (x → −2⁺) f(x) = −∞.
• x → −2⁻: x + 2 is small and negative. Negative ÷ small negative → lim (x → −2⁻) f(x) = +∞.

(c) Because lim (x → −2⁺) f(x) = −∞, the line x = −2 is a vertical asymptote of the graph of f. (Using the left-hand limit, +∞, is equally valid.)

(d) At x = 2 the limit exists and equals 3/2, so defining f(2) = 3/2 fills the hole (Topic 1.13). At x = −2 the function is unbounded, so the limit does not exist as a number. No single value of f(−2) can make the limit equal the function value, so that discontinuity cannot be removed.

Acceptable alternative for (b): a short sign chart for 2(x + 1)/(x + 2) on each side of −2. Values such as f(−1.99) = −198 and f(−2.01) = 202 can support, but on their own do not earn the point.

(a) The bottom is 0 at x = 2. For the factor (x − 2) to cancel, the top must also be 0 there (factor theorem):

2² + 2k + 6 = 0, so 10 + 2k = 0 and k = −5.

Then x² − 5x + 6 = (x − 2)(x − 3), so f(x) = x − 3 for x ≠ 2, and lim (x → 2) f(x) = 2 − 3 = −1. The graph has a hole at (2, −1), not an asymptote.

(b) With k = 0, the top tends to 4 + 6 = 10, which is positive. For x < 2, x − 2 < 0, so lim (x → 2⁻) f(x) = −∞. For x > 2, x − 2 > 0, so lim (x → 2⁺) f(x) = +∞.

(c) For x → 2⁺ the bottom is small and positive. To get −∞, the top must tend to a negative number: 10 + 2k < 0, so k < −5.

Check: k = −5 gives a hole (finite limit −1), not −∞. For k > −5 the top tends to a positive number and the right-hand limit is +∞. For example, k = −7 gives top → −4 and right-hand limit −∞, as predicted.

Acceptable alternative for (a): polynomial division of x² + kx + 6 by x − 2 gives remainder 10 + 2k; setting the remainder to 0 gives the same k. An answer of k ≤ −5 in (c) loses the last point, because k = −5 gives a finite limit.