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Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

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9 Diagnostic Tests 308 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #241 : Limits

Evaluate the limit:

$\lim_{n \to 3}\frac{2n^2-2n-12}{4n^2-8n-12}$

Possible Answers:

DNE

$\frac{5}{8}$

$-\infty$

$\infty$

Correct answer:

$\frac{5}{8}$

Explanation:

The limiting situation in this equation would be the denominator. Plug the value that n is approaching into the denominator to see if the denominator will equal 0. In this question, the denominator will equal zero when n=3; so we try to eliminate the denominator by factoring. When the denominator is no longer zero, we may continue to insert the value of n into the remaining equation.

$\lim_{n \to 3}\frac{2n^2-2n-12}{4n^2-8n-12}=\lim_{n \to 3}\frac{2(n^2-n-6)}{4(n^2-2n-3)}=\lim_{n \to 3}\frac{2(n-3)(n+2)}{4(n-3)(n+1)}=\lim_{n \to 3}\frac{n+2}{2(n+1)}=\frac{3+2}{2(3+1)}=\frac{5}{2(4)}=\frac{5}{8}$

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Example Question #242 : Limits

Evaluate the limit:

$\lim_{n \to 3}30n^2-\sqrt{n^4}$

Possible Answers:

$-\infty$

DNE

261

$\infty$

Correct answer:

261

Explanation:

There is no limiting situation in this equation (like a denominator) so we can just plug in the value that n approaches into the limit and solve:

$\lim_{n \to 3}30n^2-\sqrt{n^4}=30(3)^2-\sqrt{(3)^4}=30*9-\sqrt{81}=270-9=261$

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Example Question #243 : Limits

Evaluate the limit:

$\lim_{n \to -1}\frac{n^2+3n+2}{n^2-n-2}$

Possible Answers:

$\infty$

$-\infty$

$-\frac{1}{3}$

DNE

Correct answer:

$-\frac{1}{3}$

Explanation:

The limiting situation in this equation would be the denominator. Plug the value that n is approaching into the denominator to see if the denominator will equal 0. In this question, the denominator will equal zero when n=-1; so we try to eliminate the denominator by factoring. When the denominator is no longer zero, we may continue to insert the value of n into the remaining equation.

$\\ \lim_{n \to -1}\frac{n^2+3n+2}{n^2-n-2} \\ \\ =\lim_{n \to -1}\frac{(n+1)(n+2)}{(n+1)(n-2)} \\ \\ =\lim_{n \to -1}\frac{n+2}{n-2}=\frac{-1+2}{-1-2}=\frac{1}{-3}=-\frac{1}{3}$

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Example Question #244 : Limits

Evaluate the limit:

$\lim_{n \to -2}\frac{n^2-8n-20}{n^2+7n+10}$

Possible Answers:

$-\infty$

DNE

$\infty$

Correct answer:

Explanation:

The limiting situation in this equation would be the denominator. Plug the value that n is approaching into the denominator to see if the denominator will equal 0. In this question, the denominator will equal zero when n=-2; so we try to eliminate the denominator by factoring. When the denominator is no longer zero, we may continue to insert the value of n into the remaining equation.

$\\ \lim_{n \to -2}\frac{n^2-8n-20}{n^2+7n+10} \\ \\ =\lim_{n \to -2}\frac{(n+2)(n-10)}{(n+2)(n+5)} \\ \\ =\lim_{n \to -2}\frac{n-10}{n+5}=\frac{-2-10}{-2+5}=\frac{-12}{3}=-4$

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Example Question #245 : Limits

Evaluate the limit:

$\lim_{n \to 4}\frac{n^2-2n-8}{n^2-3n-4}$

Possible Answers:

Does Not Exist

$\infty$

$\frac{6}{5}$

$-\infty$

Correct answer:

$\frac{6}{5}$

Explanation:

The limiting situation in this equation would be the denominator. Plug the value that is approaching into the denominator to see if the denominator will equal . In this question, the denominator will equal zero when n=4 ; so we try to eliminate the denominator by factoring. When the denominator is no longer zero, we may continue to insert the value of into the remaining equation.

$\lim_{n \to 4}\frac{n^2-2n-8}{n^2-3n-4}=\lim_{n \to 4}\frac{(n-4)(n+2)}{(n-4)(n+1)}=\lim_{n \to 4}\frac{n+2}{n+1}=\frac{4+2}{4+1}=\frac{6}{5}$

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Example Question #246 : Limits

Evaluate the limit:

$\lim_{n \to 4}n^2+2n-1$

Possible Answers:

Does Not Exist

$\infty$

$-\infty$

Correct answer:

Explanation:

There is no limiting situation in this equation (like a denominator) so we can just plug in the value that approaches into the limit and solve:

$\lim_{n \to 4}n^2+2n-1=(4)^2+2(4)-1=16+8-1=24-1=23$

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Example Question #247 : Limits

Evaluate the limit:

$\lim_{n \to -4}\frac{3n^2+12n}{n^2+5n+4}$

Possible Answers:

$\infty$

$-\infty$

Does Not Exist

Correct answer:

Explanation:

The limiting situation in this equation would be the denominator. Plug the value that is approaching into the denominator to see if the denominator will equal . In this question, the denominator will equal zero when n=-4 ; so we try to eliminate the denominator by factoring. When the denominator is no longer zero, we may continue to insert the value of into the remaining equation.

$\lim_{n \to -4}\frac{3n^2+12n}{n^2+5n+4}=\lim_{n \to -4}\frac{3n(n+4)}{(n+4)(n+1)} \\ \\ =\lim_{n \to -4}\frac{3n}{n+1}=\frac{3(-4)}{-4+1}=\frac{-12}{-3}=4$

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Example Question #248 : Limits

Evaluate the limit:

$\lim_{n \to -4}5n^2+n$

Possible Answers:

$\infty$

$-\infty$

Does Not Exist

Correct answer:

Explanation:

There is no limiting situation in this equation (like a denominator) so we can just plug in the value that n approaches into the limit and solve:

$\lim_{n \to -4}5n^2+n=5(-4)^2+(-4)=5\cdot16-4=80-4=76$

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Example Question #249 : Limits

Evaluate the limit:

$\lim_{n \to 5}\frac{n^3-5n^2}{n^2-3n-10}$

Possible Answers:

$\infty$

$-\infty$

$\frac{25}{7}$

Does Not Exist

Correct answer:

$\frac{25}{7}$

Explanation:

The limiting situation in this equation would be the denominator. Plug the value that is approaching into the denominator to see if the denominator will equal . In this question, the denominator will equal zero when n=5 ; so we try to eliminate the denominator by factoring. When the denominator is no longer zero, we may continue to insert the value of into the remaining equation.

$\lim_{n \to 5}\frac{n^3-5n^2}{n^2-3n-10}=\lim_{n \to 5}\frac{n^2(n-5)}{(n-5)(n+2)}=\lim_{n \to 5}\frac{n^2}{n+2}=\frac{5^2}{5+2}=\frac{25}{7}$

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Example Question #250 : Limits

Evaluate the limit:

$\lim_{n \to 5}3n^3+2n^2+n+5$

Possible Answers:

435

Does Not Exist

$\infty$

$-\infty$

Correct answer:

435

Explanation:

There is no limiting situation in this equation (like a denominator) so we can just plug in the value that approaches into the limit and solve:

$\lim_{n \to 5}3n^3+2n^2+n+5=3(5)^3+2(5)^2+5+5 \\ \\ =3\cdot125+2\cdot25+10=375+50+10 \\ =435$

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Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #241 : Limits

Example Question #242 : Limits

Example Question #243 : Limits

Example Question #244 : Limits

Example Question #245 : Limits

Example Question #246 : Limits

Example Question #247 : Limits

Example Question #248 : Limits

Example Question #249 : Limits

Example Question #250 : Limits

All Calculus 2 Resources

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