AP Calculus BC Practice Quiz: Estimating Limit Values from Graphs

Correct Answer: B

To estimate the limit as x approaches -2, we examine the y-value the function approaches from both the left and the right of x = -2. From both sides, the graph of f(x) approaches a y-value of 1. Therefore, the limit is 1. This question assesses the ability to estimate a limit at a continuous point on a graph. [cite: 1166, 1171]

Correct Answer: B

The limit of a function as x approaches a value is the y-value the function gets closer to, not necessarily the value of the function at that point. As x approaches 3 from both the left and the right, the graph of g(x) approaches the y-value of 4. The open circle at (3, 4) indicates that g(3) is not 4, but this does not affect the limit. [cite: 1166, 1171]

Correct Answer: A

The notation x→1⁻ indicates the limit as x approaches 1 from the left side (values less than 1). Following the graph of h(x) from the left towards x=1, the function's y-values approach -2. This question specifically tests the concept of a one-sided limit. [cite: 1170, 1171]

Correct Answer: C

The notation x→1⁺ indicates the limit as x approaches 1 from the right side (values greater than 1). Following the graph of h(x) from the right towards x=1, the function's y-values approach 3. This question reinforces the concept of a one-sided limit. [cite: 1170, 1171]

Correct Answer: D

For a two-sided limit to exist, the left-sided limit must equal the right-sided limit. From the graph, as x approaches 2 from the left, f(x) approaches 1. As x approaches 2 from the right, f(x) approaches 4. Since lim (x→2⁻) f(x) ≠ lim (x→2⁺) f(x), the two-sided limit does not exist. [cite: 1170, 1175]

Correct Answer: C

Graphical representations are powerful tools, but they can be misleading. Because of issues of scale, a standard viewing window on a calculator might not reveal a very small hole, a jump, or extremely rapid oscillation near a point. This can lead to an incorrect estimation of a limit. [cite: 1173]

Correct Answer: D

For a limit to exist, the function must approach a single, finite value. In this case, the one-sided limits are different (one is +∞, the other is -∞). Since the function does not approach a single finite value, the limit does not exist. [cite: 1175]

Correct Answer: D

By examining the graph: A) As x approaches -3 from the right, the y-value approaches 2, so this is true. B) As x approaches 0 from both sides, the y-value approaches 1, so this is true. C) At x=2, there is a jump discontinuity (left limit is -1, right limit is 4), so the limit does not exist; this is true. D) At x=4, there is a hole at y=1, so the limit as x approaches 4 is 1, not 3. The value f(4) is 3, but the limit is 1. Therefore, this statement is false. [cite: 1166, 1170, 1171, 1175]

Correct Answer: C

We need to check if the left- and right-sided limits are equal at c=-2 and c=1. At c=-2, there is a hole, but the graph approaches y=3 from both the left and the right, so the limit exists. At c=1, the function is continuous, and the graph approaches y=0 from both sides, so the limit exists. Therefore, the limit exists for both values. [cite: 1166, 1171, 1175]

Correct Answer: B

First, we must find the limit and the function value separately from the graph. To find lim (x→1) f(x), we see that as x approaches 1 from both sides, the graph approaches the y-value of 4. So, the limit is 4. To find f(1), we look for the closed circle at x=1, which is at the point (1, 2). So, f(1) = 2. The expression is lim (x→1) f(x) - f(1) = 4 - 2 = 2. This question highlights the difference between the limit at a point and the function's value at that point. [cite: 1166, 1171]

Correct Answer: C

A limit does not exist at a point if the left-sided limit does not equal the right-sided limit. At x=-3, there is a hole, but the limit exists. At x=0, the function is continuous, so the limit exists. At x=2, there is a jump discontinuity; the limit from the left is different from the limit from the right. At x=4, the function is continuous. Therefore, the limit does not exist at x=2. [cite: 1175]