AP Calculus BC Quiz: Defining Limits And Using Limit Notation

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Defining Limits And Using Limit NotationQuestion 1 of 20

For $u(x)=\frac{\sin(x)}{x}$ for $x\ne0$ and $u(0)=3$ , which limit expression describes $u(x)$ as $x\to0$ ?

\displaystyle \lim_{x\to 0} u(x)=3

\displaystyle u(0)=1

\displaystyle \lim_{x\to 1} u(x)=0

\displaystyle \lim_{x\to 0} u(x)=1

\displaystyle \lim_{x\to 0} u(0)=3

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AP Calculus BC Quiz

AP Calculus BC Quiz: Defining Limits And Using Limit Notation

Practice Defining Limits And Using Limit Notation in AP Calculus BC with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Defining Limits And Using Limit Notation, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Calculus BC.

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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.

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Question 1

For $u(x)=\frac{\sin(x)}{x}$ for $x\ne0$ and $u(0)=3$ , which limit expression describes $u(x)$ as $x\to0$ ?

$\displaystyle \lim_{x\to 0} u(x)=3$
$\displaystyle u(0)=1$
$\displaystyle \lim_{x\to 1} u(x)=0$
$\displaystyle \lim_{x\to 0} u(x)=1$ (correct answer)
$\displaystyle \lim_{x\to 0} u(0)=3$

Explanation: This problem tests defining limits and using limit notation for functions with known limit properties like sin(x)/x. The expression $\lim_{x \to 0} u(x) = 1$ correctly states the well-known limit of sin(x)/x approaching 1 as x nears 0. Despite u(0)=3, the notation focuses on values around x=0, not at it. This exemplifies standard limits overriding point definitions. A tempting distractor like $\lim_{x \to 0} u(x) = 3$ fails because it confuses the redefined function value with the actual limit. For limit notation, recall standard limits, specify approach to the point, equate to the known value, and separate from f(a).

Question 2

For $h(x)=\frac{\ln(x)}{x-1}$ for $x \neq 1$ , which expression represents $\lim_{x\to1}h(x)$ ?

$\lim_{x\to1} h(x)=0$
$h(1)=1$
$\lim_{x\to1} h(x)=1$ (correct answer)
$\lim_{x=1} h(x)=1$
$\lim_{x\to0} h(x)=1$

Explanation: This question tests the skill of defining limits and using limit notation. The expression in choice C, $\lim_{x\to1} h(x)=1$ , is valid because the limit of $\ln(x)/(x-1)$ as $x$ approaches 1 is 1, a standard form. It's $0/0$ indeterminate, resolved by L'Hôpital to $1/x$ over 1, approaching 1. Both sides agree. A tempting distractor is choice A, $\lim_{x\to1} h(x)=0$ , from numerator but ignoring derivative. Use $\to$ and apply rules for indeterminates.

Question 3

Let $s(x)=\frac{\sin(3x)}{x}$ for $x\ne0$ and $s(0)=0$ ; which limit expression matches $s(x)$ as $x\to0$ ?

$\lim_{x\to0} s(x)=0$
$s(0)=3$
$\lim_{x\to0} s(x)=3$ (correct answer)
$\lim_{x\to3} s(x)=0$
$\lim_{x\to0} s(x)$ does not exist

Explanation: This question tests limit notation for trigonometric limits. The limit as x approaches 0 of s(x) is 3 because the expression simplifies to 3 times sin(3x)/(3x), which approaches 3*1=3 using the standard limit of sin(u)/u as u approaches 0. The function value s(0)=0 does not affect the limit, which examines values near 0. Therefore, the notation $\lim_{x\to0} s(x)=3$ is correct. A tempting distractor is A, which mistakenly uses the function value at x=0 for the limit. A checklist for limit notation includes specifying the function, the variable approaching a value, checking one-sided limits agree, and stating the limit value or that it does not exist.

Question 4

For $n(x)=\frac{\sqrt{x+1}-1}{x}$ for $x\ne0$ , which expression represents $\lim_{x\to0}n(x)$ ?

$\lim_{x\to0} n(x)=0$
$\lim_{x\to0} n(x)=\frac12$ (correct answer)
$n(0)=\frac12$
$\lim_{x=0} n(x)=\frac12$
$\lim_{x\to\frac12} n(x)=0$

Explanation: This question tests the skill of defining limits and using limit notation. The expression \lim_{x\to0} n(x)=\frac{1}{2} is correct because rationalizing yields ( $\sqrt{x+1}$ -1)/x = 1/( $\sqrt{x+1}$ +1) → 1/2. Both sides approach 1/2. This notation equals the limit value. A tempting distractor is C, n(0)=\frac{1}{2}, but n(0) undefined. To check limit notation, rationalize expressions, use →, and compute the value.

Question 5

For $f(x)=\begin{cases}3x,&x<2\\7,&x=2\\x^2+1,&x>2\end{cases}$ , which expression represents $\lim_{x\to2}f(x)$ ?

$\lim_{x\to2} f(x)=7$
$f(2)=7$
$\lim_{x\to2} f(x)$ does not exist (correct answer)
$\lim_{x=2} f(x)$ does not exist
$\lim_{x\to7} f(x)=2$

Explanation: This question tests the skill of defining limits and using limit notation. The expression in choice C, $\lim_{x\to2} f(x)$ does not exist, is valid because the left limit is 6 and the right is 5, differing. Left uses $3x$ approaching 6, right $x^2+1$ approaching 5. Mismatch means no two-sided limit. A tempting distractor is choice A, $\lim_{x\to2} f(x)=7$ , confusing with the point value. Use $\to$ and evaluate one-sided for piecewise.

Question 6

For $q(x)=\begin{cases}\frac{x^2-1}{x-1},&x\ne1\\0,&x=1\end{cases}$ , which expression represents $\lim_{x\to1}q(x)$ ?

$\lim_{x\to1} q(x)=0$
$q(1)=0$
$\lim_{x\to1} q(x)=2$ (correct answer)
$\lim_{x=1} q(x)=2$
$\lim_{x\to2} q(x)=1$

Explanation: This question tests the skill of defining limits and using limit notation. The expression \lim_{x\to1} q(x)=2 is correct because simplifying to x+1 for x≠1 approaches 2. The redefinition at 1=0 doesn't affect limit. This notation denotes the approached value. A tempting distractor is B, q(1)=0, but that's the value, not limit. To check limit notation, factor rationals, use →, and ignore point value.

Question 7

For $f(x)=\frac{x^2-9}{x-3}$ when $x\ne3$ and $f(3)=1$ , which expression represents the limit as $x$ approaches $3$ ?

$\lim_{x\to3} f(x)=1$
$\lim_{x\to3} f(x)=6$ (correct answer)
$f(3)=6$
$\lim_{x=3} f(x)=6$
$\lim_{x\to6} f(x)=3$

Explanation: This question tests the skill of defining limits and using limit notation. The expression in choice B, $\lim_{x\to3} f(x)=6$ , is valid because simplifying the function gives $f(x) = x+3$ for $x\ne3$ , which approaches 6 as $x$ approaches 3 from both sides. The limit focuses on the behavior near $x=3$ , not the value at $x=3$ , which is defined as 1. Even though there is a hole at $x=3$ , the function values get arbitrarily close to 6 as $x$ gets close to 3. A tempting distractor is choice A, $\lim_{x\to3} f(x)=1$ , which confuses the limit with the function value at the point. Always check the limit by simplifying expressions and evaluating the approach, and use the $\to$ symbol in the subscript for proper notation.

Question 8

For $p(x)=\begin{cases}x+4,&x<0\\4-x,&x\ge0\end{cases}$ , which expression represents $\lim_{x\to0}p(x)$ ?

$\lim_{x\to0} p(x)=0$
$p(0)=0$
$\lim_{x\to0} p(x)=4$ (correct answer)
$\lim_{x=0} p(x)=4$
$\lim_{x\to4} p(x)=0$

Explanation: This question tests the skill of defining limits and using limit notation. The expression \lim_{x\to0} p(x)=4 correctly represents the limit because from the left, x+4 approaches 4, and from the right, 4-x approaches 4. Both pieces agree at 4 as x nears 0, even though the function is defined differently on each side. This notation indicates the common value approached. A tempting distractor is B, p(0)=0, but p(0)=4-0=4, and it confuses function value with limit. To check limit notation, ensure the approach is denoted by →, the limit value is consistent from both sides, and proper syntax is used.

Question 9

For $n(x)=\frac{1}{x}$ , which expression represents the behavior as $x$ approaches $0$ ?

$\lim_{x\to0} n(x)=0$
$n(0)=0$
$\lim_{x\to0} n(x)$ does not exist (correct answer)
$\lim_{x=0} n(x)$ does not exist
$\lim_{x\to1} n(x)$ does not exist

Explanation: This question tests the skill of defining limits and using limit notation. The expression in choice C, $\lim_{x\to0} n(x)$ does not exist, is valid because as $x$ approaches 0 from the left, $n(x)$ goes to $-infty$ , and from the right to $+infty$ . The one-sided limits do not agree on a finite value. The behavior diverges in opposite directions. A tempting distractor is choice A, $\lim_{x\to0} n(x)=0$ , perhaps thinking of the pole but ignoring infinity. Use $\to$ and state 'does not exist' for divergent cases.

Question 10

For $h(x)=\frac{x^2-9}{x-3}$ when $x\ne3$ and $h(3)=10$ , which limit expression describes $h(x)$ as $x\to3$ ?

$\displaystyle h(3)=6$
$\displaystyle \lim_{x\to 3} h(x)=6$ (correct answer)
$\displaystyle \lim_{x\to 10} h(x)=3$
$\displaystyle \lim_{x\to 3} h(x)=10$
$\displaystyle \lim_{x\to 3} h(3)=10$

Explanation: This problem tests defining limits and using limit notation for algebraic functions with potential discontinuities. The expression $\lim_{x \to 3} h(x) = 6$ correctly describes the behavior as x approaches 3, since simplifying $\frac{x^2-9}{x-3}$ yields $x+3$ , which nears 6. Even though $h(3)=10$ is defined differently, the limit notation captures the nearby values, not the point itself. This illustrates removable discontinuities where limits exist despite redefinition. A tempting choice like $\lim_{x \to 3} h(x) = 10$ fails because it incorrectly uses the function value at x=3 instead of the approached value. Remember in limit notation to specify the variable and approach point, equate to the value from surrounding points, simplify expressions if needed, and separate from f(a).

Question 11

Let $r(x)=\begin{cases}x+1,&x<0\\2,&x=0\\1-x,&x>0\end{cases}$ ; which limit expression matches $r(x)$ as $x\to0$ ?

$\displaystyle \lim_{x\to 0} r(x)=2$
$\displaystyle r(0)=0$
$\displaystyle \lim_{x\to 0} r(x)=1$ (correct answer)
$\displaystyle \lim_{x\to 1} r(x)=0$
$\displaystyle \lim_{x\to 0} r(0)=2$

Explanation: This question examines defining limits and using limit notation for piecewise-defined functions. The expression $\lim_{x \to 0} r(x) = 1$ is correct because from the left, x+1 approaches 1, and from the right, 1-x approaches 1 as x nears 0. Although r(0)=2, the limit notation describes the agreement in approaching 1, not the value at 0. This demonstrates limits at points of redefinition in piecewise functions. A misleading choice like $\lim_{x \to 0} r(x) = 2$ fails as it uses r(0) instead of the nearby behavior. For limit notation, ensure left and right limits match, specify the approach point, equate to that common value, and differentiate from f(a).

Question 12

For $k(x)=\frac{2x^2-x-1}{x-1}$ when $x\ne1$ , which expression represents $\lim_{x\to1}k(x)$ ?

$\lim_{x\to1} k(x)=0$
$\lim_{x\to1} k(x)=3$ (correct answer)
$k(1)=3$
$\lim_{x=1} k(x)=3$
$\lim_{x\to3} k(x)=1$

Explanation: This question tests the skill of defining limits and using limit notation. The expression in choice B, $\lim_{x\to1} k(x)=3$ , is valid because simplifying gives $2x+1$ approaching 3 as $x$ approaches 1. Factoring removes the discontinuity. Both sides approach 3. A tempting distractor is choice C, $k(1)=3$ , but it's undefined there. Use $\to$ and simplify for removable discontinuities.

Question 13

For $t(x)=\begin{cases}x+3,&x\le0\\3-x,&x>0\end{cases}$ , which limit expression represents $t(x)$ as $x\to0$ ?

$\lim_{x\to0} t(x)=0$
$t(0)=0$
$\lim_{x\to0} t(x)=3$ (correct answer)
$\lim_{x\to3} t(x)=0$
$\lim_{x\to0} t(x)$ does not exist

Explanation: This question tests limit notation for piecewise functions. The limit as x approaches 0 of t(x) is 3 because the left-hand limit from x+3 approaches 3 and the right-hand limit from 3-x also approaches 3. The function value at x=0 is 3 from the left piece, but the limit is independent of this. Therefore, the notation $\lim_{x\to0} t(x)=3$ accurately represents the behavior. A tempting distractor is B, which states t(0)=0, perhaps from misreading the piecewise definition. A checklist for limit notation includes specifying the function, the variable approaching a value, checking one-sided limits agree, and stating the limit value or that it does not exist.

Question 14

Suppose $\lim_{x\to5^-} q(x)=1$ and $\lim_{x\to5^+} q(x)=4$ ; which limit expression correctly describes $q(x)$ as $x\to5$ ?

$\lim_{x\to5} q(x)=1$
$q(5)=4$
$\lim_{x\to5} q(x)=4$
$\lim_{x\to5} q(x)$ does not exist (correct answer)
$\lim_{x\to1} q(x)=5$

Explanation: This question tests limit notation when one-sided limits differ. The limit as x approaches 5 of q(x) does not exist because the left-hand limit is 1 and the right-hand limit is 4, which do not match. The value of q(5), if defined, is irrelevant to whether the overall limit exists. Therefore, the expression $\lim_{x\to5} q(x)$ does not exist correctly describes the situation. A tempting distractor is C, which might incorrectly take the limit as one of the one-sided values. A checklist for limit notation includes specifying the function, the variable approaching a value, checking one-sided limits agree, and stating the limit value or that it does not exist.

Question 15

For $j(x)=\frac{\tan x}{x}$ , which expression represents the behavior as $x$ approaches $0$ ?

$\lim_{x\to0} j(x)=0$
$j(0)=1$
$\lim_{x\to0} j(x)=1$ (correct answer)
$\lim_{x=0} j(x)=1$
$\lim_{x\to1} j(x)=0$

Explanation: This question tests the skill of defining limits and using limit notation. The expression \lim_{x\to0} j(x)=1 correctly represents the behavior because tan x / x approaches 1 as x nears 0, a standard limit. Values from both sides confirm this. This notation uses the arrow and equals 1. A tempting distractor is B, j(0)=1, but j(0) is undefined (0/0). To check limit notation, recall standard limits, evaluate indeterminate forms, and ensure proper syntax.

Question 16

For $u(x)=\frac{x^2-16}{x-4}$ when $x\ne4$ and $u(4)=10$ , which expression represents $\lim_{x\to4}u(x)$ ?

$\lim_{x\to4} u(x)=8$ (correct answer)
$u(4)=8$
$\lim_{x\to4} u(x)=10$
$\lim_{x=4} u(x)=8$
$\lim_{x\to8} u(x)=4$

Explanation: This question tests the skill of defining limits and using limit notation. The expression in choice A, $\lim_{x\to4} u(x)=8$ , is valid because simplifying gives $u(x)=x+4$ for $x\ne4$ , approaching 8 as $x$ approaches 4. The limit ignores the value at $x=4$ , which is 10, focusing on nearby values. Both sides approach 8 consistently. A tempting distractor is choice B, $u(4)=8$ , which uses the redefined value but not the limit behavior. Always simplify rational functions for limits and use $\to$ for proper notation.

Question 17

For $\ell(x)=\frac{x^2}{|x|}$ when $x\ne0$ , which expression represents $\lim_{x\to0}\ell(x)$ ?

$\lim_{x\to0} \ell(x)=0$ (correct answer)
$\ell(0)=0$
$\lim_{x\to0} \ell(x)=1$
$\lim_{x=0} \ell(x)=0$
$\lim_{x\to1} \ell(x)=0$

Explanation: This question tests the skill of defining limits and using limit notation. The expression \lim_{x\to0} \ell(x)=0 is correct because \ell(x)=|x| for x≠0, which approaches 0 from both sides. Left and right agree at 0. This notation denotes the limit value. A tempting distractor is B, \ell(0)=0, but \ell(0) is undefined in the given. To check limit notation, simplify absolute values, confirm one-sided limits, and use → properly.

Question 18

For $m(x)=\frac{x^2+2x-8}{x-2}$ when $x\ne2$ and $m(2)=0$ , which limit expression represents $m(x)$ as $x\to2$ ?

$\lim_{x\to2} m(x)=0$
$\lim_{x\to2} m(x)=6$ (correct answer)
$m(2)=6$
$\lim_{x\to6} m(x)=2$
$\lim_{x\to2} m(x)=2$

Explanation: This question tests the understanding of limit notation for functions with discontinuities. The limit as x approaches 2 of m(x) is 6 because the function simplifies to x+4 for x not equal to 2, which approaches 6 as x nears 2. The defined value at x=2 is 0, but the limit focuses on nearby values, not at 2 itself. Therefore, the notation $\lim_{x\to2} m(x)=6$ accurately represents this approaching behavior. A tempting distractor is A, which incorrectly equates the limit to the function value at x=2. A checklist for limit notation includes specifying the function, the variable approaching a value, checking one-sided limits agree, and stating the limit value or that it does not exist.

Question 19

Let $s(x)=\dfrac{x^2-4x}{x}$ for $x\ne0$ and $s(0)=10$ ; which statement gives $\lim_{x\to0} s(x)$ ?

$\lim_{x\to0} s(x)=10$
$\lim_{x\to0} s(x)=-4$ (correct answer)
$s(0)=-4$
$\lim_{x\to-4} s(x)=0$
$\lim_{x\to0} s(0)=-4$

Explanation: This question tests limit notation for a rational function with a removable discontinuity at x=0. The function s(x) = (x²-4x)/x can be simplified by factoring out x from the numerator: s(x) = x(x-4)/x = x-4 for x≠0. As x approaches 0, the simplified expression approaches 0-4 = -4, so lim[x→0] s(x) = -4. The given value s(0) = 10 doesn't affect the limit, which depends only on the behavior of s(x) for x near but not equal to 0. Choice E uses incorrect notation by placing s(0) inside the limit expression. For rational functions, always simplify first when possible, evaluate the limit using the simplified form, use proper lim[x→a] f(x) notation, and remember that limits are determined by nearby values, not the value at the point.

Question 20

For $q(x)=\frac{\sin(5x)}{x}$ , which limit expression correctly represents the behavior as $x\to 0$ ?

$\lim_{x\to 0} q(x)=0$
$q(0)=5$
$\lim_{x\to 0} q(x)=5$ (correct answer)
$\lim_{x\to 5} q(x)=0$
$\lim_{x\to 0} q(0)$

Explanation: This question tests limit notation for the important function $q(x) = \frac{\sin(5x)}{x}$ as $x \to 0$ . Using the standard limit $\lim_{x\to 0} \frac{\sin(x)}{x} = 1$ and substitution, we get $\lim_{x\to 0} \frac{\sin(5x)}{x} = \lim_{x\to 0} \frac{\sin(5x)}{5x} \cdot 5 = 1 \cdot 5 = 5$ . Therefore, $\lim_{x\to 0} q(x) = 5$ is the correct notation. Choice E is incomplete notation (missing the limit value) and improperly suggests evaluating at 0. For trigonometric limits: apply known limit formulas, use proper $\lim_{x\to a} q(x)$ notation, and remember that these limits describe behavior near points where direct evaluation may be undefined.