AP Calculus BC Quiz: Connecting Multiple Representations Of Limits

What this quiz covers

This quiz focuses on Connecting Multiple Representations Of Limits, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Calculus BC.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. $\lim_{x \to 1} h(x) = 4$ and $h(1)$ is undefined. (correct answer)
2. $\lim_{x \to 1} h(x)$ does not exist and $h(1)$ is undefined.
3. $\lim_{x \to 1} h(x) = 4$ and $h(1) = 4$.
4. $\lim_{x \to 4} h(x) = 1$ and $h(1)$ is undefined.

Explanation: A hole in the graph at $(1, 4)$ means two things. First, the function approaches the y-value of 4 as $x$ approaches 1 from both sides. This is represented by $\lim_{x \to 1} h(x) = 4$. Second, the function is not defined at $x=1$, so $h(1)$ is undefined. Option (A) correctly states both conditions. (B) is incorrect because the limit does exist; a hole implies the left and right limits are equal. (C) describes a function that is continuous at $x=1$, which contradicts the presence of a hole. (D) incorrectly reverses the roles of $x$ and $y$ in the limit statement.

Question 2

1. $\int_0^\pi \sin(x) dx$ (correct answer)
2. $\int_0^1 \sin(\pi x) dx$
3. $\int_0^\pi x \sin(x) dx$
4. $\int_0^1 \sin(x) dx$

Explanation: The limit of a right Riemann sum is given by $\lim_{n \to \infty} \sum_{k=1}^{n} f(a+k\Delta x) \Delta x = \int_a^b f(x) dx$, where $\Delta x = \frac{b-a}{n}$. By matching the given expression, we can identify $\Delta x = \frac{\pi}{n}$ and $f(x_k) = \sin(\frac{\pi k}{n})$. Let's choose the starting point $a=0$. Then $b-a = n \Delta x = n(\frac{\pi}{n}) = \pi$, so $b=\pi$. The sample points are $x_k = a+k\Delta x = 0 + k\frac{\pi}{n} = \frac{\pi k}{n}$. The function is $f(x) = \sin(x)$. Thus, the definite integral is $\int_0^\pi \sin(x) dx$. (B) is incorrect. The Riemann sum for this integral would be $\lim_{n \to \infty} \sum_{k=1}^{n} \sin(\frac{\pi k}{n}) \frac{1}{n}$, which is missing the factor of $\pi$ in the term corresponding to $\Delta x$. (C) is incorrect because the function in the integrand is $x \sin(x)$, which does not match the sum. (D) is incorrect because the interval of integration is $[0,1]$, which does not match the given sum.

Question 3

1. $\lim_{x \to 2^+} f(x) = 5$
2. $\lim_{x \to 2^-} f(x) = 5$ (correct answer)
3. $\lim_{x \to 5} f(x) = 2$
4. $f(2) = 5$

Explanation: The correct answer translates the verbal description into limit notation. 'Approaches 2 from values less than 2' is represented by $x \to 2^-$. 'Values that get arbitrarily close to 5' is represented by the limit equaling 5. Therefore, the correct statement is $\lim_{x \to 2^-} f(x) = 5$. (A) is incorrect because $x \to 2^+$ represents approaching 2 from values greater than 2. (C) is incorrect because it reverses the roles of the input variable and the output value of the function. (D) is incorrect because the concept of a limit describes the behavior of a function near a point, not necessarily at the point. The value of $f(2)$ could be different from 5 or even undefined.

Question 4

1. $f(x) = e^x$ and $a=2$
2. $f(x) = e^{2x}$ and $a=1$ (correct answer)
3. $f(x) = e^{x+2}$ and $a=0$
4. $f(x) = 2e^x$ and $a=1$

Explanation: The given limit matches the definition of the derivative, $f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}$. By comparing the terms, we can identify $f(a+h) = e^{2(1+h)}$ and $f(a) = e^2$. From $f(a) = e^2$, if we assume $f(x)=e^{2x}$, then $e^{2a}=e^2$, which implies $a=1$. We can verify: $f(a+h) = f(1+h) = e^{2(1+h)}$, which matches the numerator. (A) is incorrect. If $f(x)=e^x$ and $a=2$, the limit would be $\lim_{h \to 0} \frac{e^{2+h} - e^2}{h}$. (C) is incorrect. If $f(x)=e^{x+2}$ and $a=0$, then $f(a) = e^2$ but $f(a+h) = e^{h+2} \neq e^{2(1+h)}$. (D) is incorrect. If $f(x)=2e^x$ and $a=1$, then $f(a) = 2e \neq e^2$.

Question 5

1. The graphs of both $f$ and $g$ have horizontal asymptotes at $y=0$ as $x$ approaches $a$.
2. The graphs of both $f$ and $g$ must pass through the origin at the point $(0,0)$.
3. If $f$ and $g$ are continuous at $a$, then the graphs of both functions have an x-intercept at $x=a$. (correct answer)
4. The graphs of both $f$ and $g$ must have local extrema (a minimum or maximum) at $x=a$.

Explanation: The condition $\lim_{x \to a} f(x) = 0$ means that the function values of $f$ approach 0 as $x$ approaches $a$. If $f$ is also continuous at $x=a$, this implies that $f(a)=0$. Graphically, a point where a function's value is zero is an x-intercept. Therefore, for continuous functions, both graphs would have an x-intercept at $x=a$. This is the most direct graphical interpretation. (A) is incorrect. Horizontal asymptotes relate to limits as $x \to \pm\infty$, not as $x \to a$. (B) is only true if $a=0$. The statement must hold for any value of $a$. (D) is not necessarily true. Having a root at $x=a$ does not guarantee a local extremum; for example, $f(x)=(x-a)^3$ has a root but no extremum at $x=a$.

Question 6

1. Because $\lim_{x \to 0} (-x^2) = 0$ and $\lim_{x \to 0} x^2 = 0$, it follows that $\lim_{x \to 0} x^2 \cos(\frac{\pi}{x}) = 0$. (correct answer)
2. Because $\lim_{x \to 0} x^2=0$, the entire expression must go to 0, which is confirmed by the theorem.
3. Because $\cos(\frac{\pi}{x})$ oscillates between -1 and 1, the limit does not exist, so the theorem is not applicable.
4. Because $\lim_{x \to 0} (-x^2) \neq \lim_{x \to 0} x^2$, the Squeeze Theorem cannot be used to find the limit in this case.

Explanation: The Squeeze Theorem states that if $g(x) \le f(x) \le h(x)$ for all $x$ in an open interval containing $c$ (except possibly at $c$ itself), and if $\lim_{x \to c} g(x) = \lim_{x \to c} h(x) = L$, then $\lim_{x \to c} f(x) = L$. In this case, $g(x)=-x^2$, $h(x)=x^2$, and we find the limits of both bounding functions as $x \to 0$. Since both limits are 0, the limit of the function in the middle must also be 0. Option (A) correctly represents this logical step. (B) is an incomplete justification; the Squeeze Theorem requires showing the limits of both bounding functions are equal. (C) misunderstands the purpose of the theorem, which is specifically designed to handle cases involving oscillating functions like this one. (D) is factually incorrect, as $\lim_{x \to 0} (-x^2) = 0$ and $\lim_{x \to 0} x^2 = 0$.

Question 7

1. $\lim_{x \to -2^-} k(x) = \infty$ and $\lim_{x \to -2^+} k(x) = -\infty$ (correct answer)
2. $\lim_{x \to \infty} k(x) = -2$ and $\lim_{x \to -\infty} k(x) = -2$
3. $\lim_{x \to -2} k(x) = \infty$ and $\lim_{x \to -2} k(x) = -\infty$
4. $\lim_{x \to -2^-} k(x) = -\infty$ and $\lim_{x \to -2^+} k(x) = \infty$

Explanation: The verbal description needs to be translated into one-sided limit notation. 'Increase without bound' means the limit is $\infty$. 'Decrease without bound' means the limit is $-\infty$. 'Approaches -2 from the left' is $x \to -2^-$. 'Approaches -2 from the right' is $x \to -2^+$. Combining these gives the pair of statements in (A). (B) describes a horizontal asymptote at $y=-2$. (C) is incorrect because a two-sided limit cannot equal both $+\infty$ and $-\infty$; in this case, the two-sided limit does not exist. (D) reverses the behavior on the left and right sides of the asymptote.

Question 8

1. Limit exists and equals $2$; graph shows jump; table approaches $1$ and $3$; algebra matches $2$.
2. Limit does not exist; graph shows jump at $x=2$; table approaches $1$ from left and $3$ from right; algebra $f(x)=\begin{cases}1,&x<2\\3,&x>2\end{cases}$. (correct answer)
3. Limit equals $1$; graph shows jump; table approaches $1$ and $3$; algebra $f(x)=\begin{cases}1,&x<2\\3,&x>2\end{cases}$.
4. Limit equals $3$; graph shows jump; table approaches $1$ and $3$; algebra $f(x)=\begin{cases}1,&x<2\\3,&x>2\end{cases}$.
5. Limit equals $2$; graph has hole at $(2,2)$; table approaches $2$; algebra $f(x)=\begin{cases}1,&x<2\\3,&x>2\end{cases}$.

Explanation: Connecting multiple representations of limits is a key skill in AP Calculus BC that requires interpreting tables, graphs, and algebraic forms to determine limit values. The table showing values approaching 1 from the left and 3 from the right indicates the limit does not exist as x approaches 2. The graph with a jump at x=2 represents different one-sided limits. Algebraically, the piecewise function with 1 for x<2 and 3 for x>2 confirms the left and right limits differ. A tempting distractor might be choice A, which incorrectly claims the limit exists as 2, ignoring the jump discontinuity. To connect representations effectively, always check one-sided limits in piecewise functions or jumps for existence.

Question 9

1. Table approaches $1$ from both sides; graph approaches $1$; algebra gives limit $1$.
2. Table approaches $-1$ from both sides; graph approaches $-1$; algebra gives limit $-1$.
3. Table approaches $-1$ from left and $1$ from right; graph has jump at $x=0$; algebra shows limit does not exist. (correct answer)
4. Table approaches $0$ from both sides; graph crosses origin; algebra gives limit $0$.
5. Table diverges to $\infty$; graph has vertical asymptote at $x=0$; algebra gives limit $\infty$.

Explanation: Connecting multiple representations of limits is a key skill in AP Calculus BC that requires interpreting tables, graphs, and algebraic forms to determine limit values. The table showing values approaching -1 from the left and 1 from the right indicates the limit does not exist as x approaches 0. The graph with a jump at x=0 represents different one-sided limits. Algebraically, r(x) = |x|/x is -1 for x<0 and 1 for x>0, confirming the limit does not exist. A tempting distractor might be choice A, which incorrectly claims the table approaches 1 from both sides, overlooking the absolute value's effect. To connect representations effectively, always check one-sided limits in functions with absolute values or discontinuities.

Question 10

1. Algebra: $\lim_{x\to4}s(x)$ does not exist; Table: values go to $-\infty$ from left and $\infty$ from right; Graph: vertical asymptote at $x=4$. (correct answer)
2. Algebra: $\lim_{x\to4}s(x)=0$; Table: values go to $-\infty$ from left and $\infty$ from right; Graph: vertical asymptote at $x=4$.
3. Algebra: $\lim_{x\to4}s(x)$ does not exist; Table: values approach $0$ from both sides; Graph: vertical asymptote at $x=4$.
4. Algebra: $\lim_{x\to4}s(x)$ does not exist; Table: values go to $-\infty$ from left and $\infty$ from right; Graph: removable hole at $(4,0)$.
5. Algebra: $\lim_{x\to4}s(x)=\infty$; Table: values go to $-\infty$ from left and $\infty$ from right; Graph: vertical asymptote at $x=4$.

Explanation: This problem requires connecting multiple representations of limits for s(x) = 1/(x-4). Algebraically, as x approaches 4 from the left, x-4 approaches 0 through negative values, making s(x) approach -∞; from the right, x-4 approaches 0 through positive values, making s(x) approach +∞. A table would show s(3.9) = -10, s(3.99) = -100, s(4.1) = 10, s(4.01) = 100, confirming the one-sided limits approach opposite infinities. Graphically, this creates a vertical asymptote at x=4 with the function going to -∞ on the left and +∞ on the right. Choice B incorrectly claims the limit is 0, perhaps thinking of horizontal asymptotes instead of vertical ones. When one-sided limits approach different infinities, the two-sided limit does not exist, which all three representations must consistently show.

Question 11

1. The graph of $f(x)$ has a vertical asymptote at the line $x=7$.
2. The graph of $f(x)$ has a horizontal asymptote at the line $y=7$. (correct answer)
3. The graph of $f(x)$ has a removable discontinuity at a point where the y-coordinate is 7.
4. The graph of $f(x)$ has an x-intercept at $x=7$.

Explanation: A limit at infinity describes the end behavior of a function. The statement $\lim_{x \to \infty} f(x) = L$ means that the function values $f(x)$ get arbitrarily close to $L$ as $x$ increases without bound. Graphically, this corresponds to a horizontal asymptote at the line $y=L$. In this case, $L=7$. (A) is incorrect; a vertical asymptote at $x=7$ would be represented by a limit like $\lim_{x \to 7} f(x) = \pm\infty$. (C) is incorrect because a limit at infinity describes end behavior, not behavior at a specific point. (D) is incorrect; an x-intercept at $x=7$ means $f(7)=0$, which is unrelated to the limit at infinity.

Question 12

1. 3.98
2. 4 (correct answer)
3. 1
4. The limit does not exist.

Explanation: The provided table of values shows that as $t$ approaches 1 from both the left (with values like 0.9 and 0.99) and the right (with values like 1.01 and 1.1), the corresponding function values $h(t)$ are getting closer and closer to 4. This numerical representation strongly suggests that the limit is 4. (A) is incorrect because it is one of the function values for an input close to 1, but not the value being approached from both sides. (C) is incorrect because 1 is the value that the input variable $t$ is approaching, not the value the function is approaching. (D) is incorrect because the values from the left and the right are both approaching the same number, 4.

Question 13

1. The value of $f(a)$ exists and the limit $\lim_{x \to a} f(x)$ exists.
2. $\lim_{x \to a} f(x) = f(a)$ (correct answer)
3. The values of $\lim_{x \to a^-} f(x)$ and $\lim_{x \to a^+} f(x)$ are both finite and equal.
4. The graph of the function does not have a break or jump at the point $x=a$.

Explanation: The formal definition of continuity at a point $a$ requires three conditions to be met: (1) $f(a)$ is defined, (2) $\lim_{x \to a} f(x)$ exists, and (3) $\lim_{x \to a} f(x) = f(a)$. The single statement in option (B) concisely encapsulates all three of these conditions. The existence of both sides of the equation is implied. (A) is incomplete; it states the first two conditions but omits the crucial third condition that the limit must equal the function value. (C) is equivalent to saying the two-sided limit exists, but it still omits the condition that this limit must equal $f(a)$. (D) is a correct intuitive or verbal description, but it is not the precise mathematical definition based on limits.

Question 14

1. $\lim_{x \to 0} p(x) = -\infty$
2. $\lim_{x \to -\infty} p(x) = 0$ (correct answer)
3. $\lim_{x \to \infty} p(x) = 0$
4. $\lim_{x \to 0^-} p(x) = -\infty$

Explanation: The phrase 'as $x$ decreases without bound' means that $x$ is approaching negative infinity, which is written symbolically as $x \to -\infty$. The phrase 'the corresponding values of $p(x)$ approach 0' means the limit equals 0. Combining these gives the statement $\lim_{x \to -\infty} p(x) = 0$. (A) describes behavior near $x=0$, where function values decrease without bound. (C) describes the end behavior as $x$ increases without bound. (D) describes a one-sided limit as $x$ approaches 0 from the left.

Question 15

1. A jump discontinuity, because the function value does not exist at that point.
2. A vertical asymptote, because the function is not defined at that point.
3. A removable discontinuity, because the two-sided limit exists but is not equal to the function value. (correct answer)
4. The graph must be continuous at $x=3$, because the limit exists at that point.

Explanation: A removable discontinuity, or a hole in the graph, occurs when the limit of the function exists at a point, but the function value at that point is either different from the limit or is undefined. The given conditions, $\lim_{x \to 3} g(x) = 5$ and $g(3)$ being undefined, perfectly match this definition. The discontinuity is 'removable' because we could define $g(3)=5$ to make the function continuous. (A) is incorrect. A jump discontinuity requires the left and right-hand limits to be different. (B) is incorrect. A vertical asymptote requires one or both one-sided limits to be infinite. (D) is incorrect. For continuity, the limit must exist, the function must be defined, and they must be equal.

Question 16

1. $\lim_{b \to \infty} \int_1^b \frac{1}{x^p} dx$ (correct answer)
2. $\lim_{a \to 1^+} \int_a^\infty \frac{1}{x^p} dx$
3. $\int_1^b \frac{1}{x^p} dx$, for a very large number $b$
4. $\lim_{p \to \infty} \int_1^\infty \frac{1}{x^p} dx$

Explanation: An improper integral with an infinite upper limit of integration is defined by replacing the infinite limit with a variable (e.g., $b$) and then taking the limit as that variable approaches infinity. This transforms the problem of finding the area of an infinite region into the limit of the areas of finite regions. (B) is incorrect because the lower limit is a finite number, so there is no need to take a limit as a variable approaches it from the right. (C) is an intuitive approximation but not the formal mathematical definition, which requires the use of a limit. (D) is incorrect because $p$ is a parameter in the integrand; the integral is defined with respect to the variable of integration, not the parameter.

Question 17

1. $\lim_{n \to \infty} a_n = S$
2. $\lim_{k \to \infty} S_k = S$ (correct answer)
3. $\lim_{k \to \infty} S_k = 0$
4. $\lim_{n \to \infty} a_n = 0$

Explanation: The definition of convergence for an infinite series is based on the behavior of its sequence of partial sums. The series converges to a sum $S$ if and only if the limit of the sequence of its partial sums exists and is equal to $S$. This is correctly stated as $\lim_{k \to \infty} S_k = S$. (A) is incorrect; the limit of the terms must be 0 for the series to converge, not S. (C) is incorrect unless the sum of the series happens to be 0. (D) is the statement of the n-th Term Test for Divergence. While it is a necessary condition for convergence, it does not define what the series converges to; it only states that if the limit is not 0, the series diverges.

Question 18

1. Algebra: left limit $=-1$, right limit $=1$, so limit does not exist; Graph: jump at $x=0$; Table: approaches $-1$ and $1$. (correct answer)
2. Algebra: left limit $=-1$, right limit $=1$, so limit equals $0$; Graph: jump at $x=0$; Table: approaches $-1$ and $1$.
3. Algebra: left limit $=-1$, right limit $=-1$, so limit equals $-1$; Graph: jump at $x=0$; Table: approaches $-1$ and $1$.
4. Algebra: left limit $=-1$, right limit $=1$, so limit does not exist; Graph: open circle at $(0,0)$; Table: approaches $-1$ and $1$.
5. Algebra: left limit $=-1$, right limit $=1$, so limit does not exist; Graph: continuous through $(0,1)$; Table: approaches $-1$ and $1$.

Explanation: Connecting multiple representations of limits involves understanding how algebraic, graphical, and tabular forms convey the same limiting behavior. Algebraically, for q(x) = |x|/x, the left-hand limit is -1 and the right-hand limit is 1, so the overall limit does not exist. Graphically, a jump discontinuity at x = 0 illustrates the differing one-sided limits. Tabular values approach -1 from the left and 1 from the right, confirming the discrepancy. Choice B fails as a tempting distractor by claiming the limit is 0, perhaps averaging the sides incorrectly. A transferable strategy is to evaluate one-sided limits algebraically, identify jumps or asymptotes graphically, and examine directional approaches in tables.

Question 19

1. Algebra: $\lim_{x\to1}u(x)=3$; Table: left approaches $3$ and right approaches $2$; Graph: open circle at $(1,3)$ from left and open circle at $(1,2)$ from right with filled dot at $(1,3)$.
2. Algebra: $\lim_{x\to1}u(x)$ does not exist; Table: left approaches $3$ and right approaches $2$; Graph: open circle at $(1,3)$ from left and open circle at $(1,2)$ from right with filled dot at $(1,3)$. (correct answer)
3. Algebra: $\lim_{x\to1}u(x)=2$; Table: left approaches $3$ and right approaches $2$; Graph: open circle at $(1,3)$ from left and open circle at $(1,2)$ from right with filled dot at $(1,3)$.
4. Algebra: $\lim_{x\to1}u(x)$ does not exist; Table: both sides approach $3$; Graph: open circle at $(1,3)$ from left and open circle at $(1,2)$ from right with filled dot at $(1,3)$.
5. Algebra: $\lim_{x\to1}u(x)$ does not exist; Table: left approaches $3$ and right approaches $2$; Graph: open circle at $(1,2)$ from left and open circle at $(1,3)$ from right with filled dot at $(1,3)$.

Explanation: This problem requires connecting multiple representations of limits for a piecewise function u(x). Algebraically, for x<1, u(x) = x+2, so the left-hand limit is 1+2 = 3; for x>1, u(x) = x²+1, so the right-hand limit is 1²+1 = 2; since these differ, lim(x→1) u(x) does not exist. A table would show u(0.9) = 2.9, u(0.99) = 2.99 from the left and u(1.1) = 2.21, u(1.01) = 2.0201 from the right, confirming different one-sided limits. Graphically, this creates open circles at (1,3) from the left piece and (1,2) from the right piece, with a filled dot at (1,3) for the actual value u(1)=3. Choice A incorrectly claims the limit exists and equals 3, considering only the left-hand limit. For piecewise functions, verify that all representations consistently show whether one-sided limits match to determine if the two-sided limit exists.

Question 20

1. Algebra: $\lim_{x\to0}h(x)=2$; Table: values approach $1$; Graph: open circle at $(0,1)$ and filled dot at $(0,2)$.
2. Algebra: $\lim_{x\to0}h(x)=1$; Table: values approach $1$; Graph: open circle at $(0,1)$ and filled dot at $(0,2)$. (correct answer)
3. Algebra: $\lim_{x\to0}h(x)$ does not exist; Table: values approach $1$; Graph: open circle at $(0,1)$ and filled dot at $(0,2)$.
4. Algebra: $\lim_{x\to0}h(x)=1$; Table: values approach $2$; Graph: open circle at $(0,1)$ and filled dot at $(0,2)$.
5. Algebra: $\lim_{x\to0}h(x)=1$; Table: values approach $1$; Graph: open circle at $(0,2)$ and filled dot at $(0,1)$.

Explanation: This problem requires connecting multiple representations of limits for h(x) = sin(x)/x, a famous limit in calculus. Algebraically, using L'Hôpital's rule or the squeeze theorem, we can show that lim(x→0) sin(x)/x = 1. A table of values confirms this: h(0.1) ≈ 0.998, h(0.01) ≈ 0.99998, h(-0.1) ≈ 0.998, h(-0.01) ≈ 0.99998, showing convergence to 1 from both sides. Graphically, this appears as an open circle at (0,1) where the limit occurs, and a filled dot at (0,2) showing the assigned value h(0)=2. Choice A incorrectly claims the limit is 2, confusing the function value h(0)=2 with the actual limit of 1. When analyzing limits, focus on the behavior as x approaches the point, not the value at the point itself.