Defining Limits and Using Limit Notation - AP Calculus BC

1. For $m(x)=\frac{x^2+2x-8}{x-2}$ when $x\ne2$ and $m(2)=0$, which limit expression represents $m(x)$ as $x\to2$?
   Answer: $\lim_{x\to2} m(x)=6$ 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.
2. For $u(x)=\frac{x^2-16}{x-4}$ when $x\ne4$ and $u(4)=10$, which expression represents $\lim_{x\to4}u(x)$?
   Answer: $\lim_{x\to4} u(x)=8$ 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.
3. For $w(x)=\begin{cases}2,&x<1\\x^2+1,&x\ge1\end{cases}$, which expression represents $\lim_{x\to1}w(x)$?
   Answer: $\lim_{x\to1} w(x)=2$ This question tests the skill of defining limits and using limit notation. The expression in choice A, $\lim_{x\to1} w(x)=2$, is valid because the left limit is 2 and the right limit is $1^2+1=2$, agreeing on 2. The piecewise definition has left at 2 and right approaching 2. The limit exists since both sides match. A tempting distractor is choice C, $\lim_{x\to1} w(x)$ does not exist, assuming mismatch but overlooking the calculation. Use $\to$ in notation and compute one-sided limits for piecewise functions.
4. For $n(x)=\frac{\sqrt{x+1}-1}{x}$ for $x\ne0$, which expression represents $\lim_{x\to0}n(x)$?
   Answer: $\lim_{x\to0} n(x)=\frac12$ 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.
5. For $h(x)=\frac{x^2-9}{x-3}$ when $x\ne3$ and $h(3)=10$, which limit expression describes $h(x)$ as $x\to3$?
   Answer: $\displaystyle \lim_{x\to 3} h(x)=6$ 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).
6. For $u(x)=\frac{\sin(x)}{x}$ for $x\ne0$ and $u(0)=3$, which limit expression describes $u(x)$ as $x\to0$?
   Answer: $\displaystyle \lim_{x\to 0} u(x)=1$ 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).
7. For $r(x)=\frac{\sin(5x)}{x}$ for $x\ne0$ and $r(0)=0$, which limit expression represents the value approached as $x\to0$?
   Answer: $\lim_{x\to0} r(x)=5$ This question involves limit notation for a trigonometric indeterminate form. The function r(x) = sin(5x)/x requires the limit identity lim[x→0] sin(x)/x = 1. Rewriting as 5·sin(5x)/(5x), we can substitute u = 5x to get 5·lim[u→0] sin(u)/u = 5·1 = 5. Therefore, lim[x→0] r(x) = 5. The assigned value r(0) = 0 doesn't affect the limit calculation. Choice A incorrectly states the limit is 0, perhaps confusing it with r(0). When evaluating trig limits, use standard limit identities and algebraic manipulation to match known forms.
8. For $t(x)=\frac{1}{x-2}$, which limit expression correctly represents the behavior as $x\to2$?
   Answer: $\lim_{x\to2} t(x)$ does not exist This question examines limit notation for a function with a vertical asymptote. The function t(x) = 1/(x-2) has a vertical asymptote at x = 2. As x approaches 2 from the right, t(x) → +∞, while as x approaches 2 from the left, t(x) → -∞. Since the one-sided limits are infinite with opposite signs, lim[x→2] t(x) does not exist. Choice A incorrectly claims the limit is 0, perhaps thinking 1/0 = 0. Choice E uses improper notation with t(2) inside the limit. For rational functions, check for vertical asymptotes where denominators equal zero; limits typically don't exist at these points.
9. For $h(x)=\frac{\ln(x)}{x-1}$ for $x \neq 1$, which expression represents $\lim_{x\to1}h(x)$?
   Answer: $\lim_{x\to1} h(x)=1$ 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.
10. A function $p$ satisfies $\lim_{x\to4^-}p(x)=1$ and $\lim_{x\to4^+}p(x)=1$ with $p(4)=7$; which limit expression is correct?
    Answer: $\lim_{x\to4} p(x)=1$ This question tests understanding of limit notation when left and right limits agree. Since lim[x→4⁻] p(x) = 1 and lim[x→4⁺] p(x) = 1, the two-sided limit exists and equals 1, so lim[x→4] p(x) = 1. The function value p(4) = 7 is independent of the limit value. Choice B incorrectly states the limit equals 7, confusing the function value at x = 4 with the limit as x approaches 4. Choice E uses improper notation with p(4) inside the limit. Remember: when one-sided limits match, write the two-sided limit; function values don't determine limit values.
11. If $s(x)=\frac{|x-5|}{x-5}$ for $x\ne 5$ and $s(5)=0$, which expression describes $\lim_{x\to 5} s(x)$?
    Answer: $\lim_{x\to 5} s(x)$ does not exist This question involves limit notation for $s(x) = \frac{|x-5|}{x-5}$, which equals 1 for $x>5$ and $-1$ for $x<5$. Since the left-hand limit is $-1$ and the right-hand limit is 1, these one-sided limits are different. Therefore, the two-sided limit does not exist, making "$\lim_{x\to 5} s(x)$ does not exist" the correct statement. The value $s(5)=0$ is irrelevant to the limit. Choice E incorrectly places $s(5)$ inside the limit notation, which is syntactically wrong. For limits of piecewise or absolute value functions: check one-sided limits separately, declare "does not exist" when they differ, and use proper notation without function evaluations inside the limit symbol.
12. Let $s(x)=\frac{\sqrt{x+9}-3}{x}$ for $x\ne0$ and $s(0)=4$; which limit expression matches $x\to0$?
    Answer: $\lim_{x\to0} s(x)=\frac{1}{6}$ This question tests limit notation for an algebraic indeterminate form. To evaluate lim[x→0] (√(x+9)-3)/x, multiply by the conjugate: [(√(x+9)-3)/x]·[(√(x+9)+3)/(√(x+9)+3)] = (x+9-9)/[x(√(x+9)+3)] = x/[x(√(x+9)+3)] = 1/(√(x+9)+3). As x→0, this becomes 1/(√9+3) = 1/6. Thus lim[x→0] s(x) = 1/6. The value s(0) = 4 is irrelevant to the limit. Choice A incorrectly uses the assigned function value. For limits involving radicals, rationalize to eliminate indeterminate forms before evaluating.
13. Given $h(x)=\begin{cases}x^2,&x\ne -2\\7,&x=-2\end{cases}$, which limit expression correctly describes $h(x)$ as $x\to -2$?
    Answer: $\lim_{x\to -2} h(x)=4$ This question tests limit notation for a piecewise function where the limit exists but differs from the function value. For $h(x) = x^2$ when $x \neq -2$ and $h(-2) = 7$, we need to find $\lim_{x\to -2} h(x)$. As $x$ approaches $-2$, the function values approach $(-2)^2 = 4$, so $\lim_{x\to -2} h(x) = 4$ is the correct notation. The fact that $h(-2) = 7$ doesn't affect the limit value, which depends only on nearby values. Choice E incorrectly includes $h(-2)$ inside the limit notation, which is syntactically wrong. Remember for limits: use $\lim_{x\to a} h(x)$ format, evaluate using the function's behavior near (not at) the point, and never place function evaluations inside limit symbols.
14. For $f(x)=\frac{x^2-9}{x-3}$ when $x\ne3$ and $f(3)=1$, which expression represents the limit as $x$ approaches $3$?
    Answer: $\lim_{x\to3} f(x)=6$ 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.
15. For $n(x)=\frac{1}{x}$, which expression represents the behavior as $x$ approaches $0$?
    Answer: $\lim_{x\to0} n(x)$ does not exist 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.
16. Let $s(x)=\frac{\sin(3x)}{x}$ for $x\ne0$ and $s(0)=0$; which limit expression matches $s(x)$ as $x\to0$?
    Answer: $\lim_{x\to0} s(x)=3$ 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.
17. For $q(x)=\frac{\sin(x)}{x}$, which expression correctly represents the behavior as $x$ approaches $0$?
    Answer: $\lim_{x\to0} q(x)=1$ This question tests the skill of defining limits and using limit notation. The expression in choice C, $\lim_{x\to0} q(x)=1$, is valid because the standard limit of $\sin(x)/x$ as $x$ approaches 0 is 1 from both sides. This holds even though the function is undefined at $x=0$. The values approach 1 as $x$ gets arbitrarily close to 0. A tempting distractor is choice A, $\lim_{x\to0} q(x)=0$, perhaps confusing with $\sin(0)=0$ but ignoring the division by $x$. Always use the $\to$ symbol in limit notation, and recall standard limits for common functions like this.
18. For $j(x)=\frac{\tan x}{x}$, which expression represents the behavior as $x$ approaches $0$?
    Answer: $\lim_{x\to0} j(x)=1$ 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.
19. Given $g(x)=\frac{|x|}{x}$ for $x\ne0$ and $g(0)=2$, which limit expression matches the behavior as $x\to0$?​
    Answer: $\lim_{x\to0} g(x)$ does not exist This question examines limit notation for a function with a jump discontinuity. The function g(x) = |x|/x equals 1 when x > 0 and -1 when x < 0, creating different left and right limits at x = 0. As x approaches 0 from the right, g(x) = 1, while as x approaches 0 from the left, g(x) = -1. Since the left and right limits differ, lim[x→0] g(x) does not exist. The value g(0) = 2 doesn't affect the limit's existence. Choice A incorrectly claims the limit equals 2, confusing the function value with the limit. For limits to exist, check that left-hand limit equals right-hand limit at the point of interest.
20. For $d(x)=\frac{x-5}{\sqrt{x}-\sqrt{5}}$ for $x\ne5$, which expression represents $\lim_{x\to5}d(x)$?
    Answer: $\lim_{x\to5} d(x)=2\sqrt{5}$ This question tests the skill of defining limits and using limit notation. The expression in choice B, $\lim_{x\to5} d(x)=2\sqrt{5}$, is valid because rationalizing gives $\sqrt{x} + \sqrt{5}$ approaching $2\sqrt{5}$. The limit ignores the undefined point at 5. Both sides approach the same value. A tempting distractor is choice A, $\lim_{x\to5} d(x)=0$, from direct substitution but missing simplification. Use $\to$ and algebraic techniques for indeterminate forms.

Start this quiz