AP Calculus BC Practice Quiz: Connecting Infinite Limits and Vertical Asymptotes
Written by AP Content Team, Verified for 2026 AP Exams, Last updated: May 2026
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A) As x approaches 2 from the right, the values of f(x) increase without bound.
B) As x approaches 2 from the left, the values of f(x) increase without bound.
C) The graph of f(x) has a hole at x = 2.
D) The value of the function at x=2 is infinity.
Correct Answer: A
The notation `lim_{x→2⁺} f(x) = ∞` specifically describes the behavior of the function as x gets closer to 2 from values greater than 2 (the right side). The limit being `∞` indicates that the function's values grow infinitely large, which is described as increasing without bound.
A) `lim_{x→-1} g(x)` exists and is finite.
B) `lim_{x→∞} g(x) = -1`.
C) At least one of the limits `lim_{x→-1⁺} g(x)` or `lim_{x→-1⁻} g(x)` is `∞` or `-∞`.
D) `g(-1)` must be defined.
Correct Answer: C
The definition of a vertical asymptote at `x = c` is that the function's values approach `∞` or `-∞` as x approaches `c` from the left, the right, or both. This means at least one of the one-sided limits must be infinite, describing the unbounded behavior of the function.
A) The limit exists and is equal to the real number infinity.
B) The function `f(x)` is defined at `x=c` and its value is `∞`.
C) The limit does not exist because the values of `f(x)` increase without bound as `x` approaches `c`.
D) The graph of `f(x)` must have a horizontal asymptote at `y=c`.
Correct Answer: C
In the formal definition of a limit, a limit only "exists" if it approaches a specific, finite real number. An infinite limit is a specific way in which a limit fails to exist. It is used to describe the unbounded behavior of a function as its values grow arbitrarily large.
A) The graph has a jump discontinuity at `x = 4`.
B) The graph has a vertical asymptote at `x = 4`, with the function decreasing without bound on the left and increasing without bound on the right.
C) The graph has a removable discontinuity (a hole) at `x = 4`.
D) The two-sided limit `lim_{x→4} h(x)` exists and is infinite.
Correct Answer: B
The two one-sided limits describe the behavior around a vertical asymptote. `lim_{x→4⁻} h(x) = -∞` means the graph goes down to negative infinity as x approaches 4 from the left. `lim_{x→4⁺} h(x) = ∞` means the graph goes up to positive infinity as x approaches 4 from the right. This is a classic example of asymptotic and unbounded behavior.
A) `lim_{x→1⁺} f(x) = ∞`
B) `lim_{x→1⁻} f(x) = -∞`
C) `lim_{x→1⁻} f(x) = ∞`
D) `lim_{x→-2} f(x) = ∞`
Correct Answer: C
The function has a vertical asymptote where the denominator is zero, at `x=1`. To evaluate the limit as `x` approaches 1 from the left (`x→1⁻`), we consider values slightly less than 1 (e.g., 0.99). The numerator is a negative constant (-2). The denominator `(x-1)` will be a small negative number. A negative constant divided by a small negative number results in a large positive number. Therefore, the limit is `∞`.
A) `lim_{x→a} f(x)` exists and is a finite number.
B) The numerator of `f(x)` is also zero at `x=a`.
C) At least one of the one-sided limits of `f(x)` as `x` approaches `a` is `∞` or `-∞`.
D) The function `f` must be continuous everywhere else.
Correct Answer: C
A function being undefined at `x=a` could indicate either a vertical asymptote or a removable discontinuity (a hole). The defining characteristic of a vertical asymptote is that the function exhibits unbounded behavior. This is formally described by one or both of the one-sided limits being infinite (`∞` or `-∞`).
A) The line `x=-2` is a vertical asymptote.
B) `lim_{x→2⁺} g(x) = -∞`
C) `lim_{x→-2} g(x) = ∞`
D) `lim_{x→2⁻} g(x) = -∞`
Correct Answer: D
First, factor the denominator: `g(x) = rac{x+2}{(x-2)(x+2)}`. The `(x+2)` term cancels, leaving `g(x) = rac{1}{x-2}` for `x ≠ -2`. This means there is a hole at `x=-2`, not a vertical asymptote. The vertical asymptote occurs where the simplified denominator is zero, at `x=2`. To find the limit as `x` approaches 2 from the left (`x→2⁻`), we consider a value slightly less than 2 (e.g., 1.99). The numerator is 1 (positive). The denominator `(x-2)` will be a small negative number. A positive number divided by a small negative number results in a large negative number. Thus, `lim_{x→2⁻} g(x) = -∞`.
A) `lim_{x→c⁻} f(x) = L`, where L is a real number.
B) `lim_{x→-∞} f(x) = c`.
C) `lim_{x→c⁻} f(x) = ∞` or `lim_{x→c⁻} f(x) = -∞`.
D) `f(c)` is not defined.
Correct Answer: C
"Unbounded behavior" means that the function's values are not approaching a finite number; instead, they are growing infinitely large (positive) or infinitely small (negative). The limit notation for this as `x` approaches `c` from the left (`x→c⁻`) is that the one-sided limit is either `∞` or `-∞`.
A) `lim_{x→1⁺} f(x) = -∞`
B) `lim_{x→1⁻} f(x) = -∞`
C) `lim_{x→-∞} f(x) = 1`
D) `lim_{x→1} f(x) = -∞`
Correct Answer: A
The phrase "As `x` approaches 1" corresponds to the limit as `x→1`. "from the right side" specifies the one-sided limit `x→1⁺`. The phrase "the y-values of the function decrease without bound" means the function's value is approaching `-∞`. Combining these pieces gives the limit statement `lim_{x→1⁺} f(x) = -∞`.