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Calculus 2 : Limits

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Example Questions

Example Question #261 : Limits

Evalute the limit:

$\lim_{n \to 3}\frac{n^2+n-12}{n^2-5n+6}$

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=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 into the remaining equation.

$\lim_{n \to 3}\frac{n^2+n-12}{n^2-5n+6}=\lim_{n \to 3}\frac{(n-3)(n+4)}{(n-3)(n-2)}=\lim_{n \to 3}\frac{n+4}{n-2}=\frac{3+4}{3-2}=\frac{7}{1}=7$

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Example Question #262 : Limits

Evaluate the limit:

$\lim_{n \to 4}\frac{n^2-16}{n^2-3n-4}$

Possible Answers:

$\infty$

$\frac{8}{5}$

$-\infty$

Does not exist

Correct answer:

$\frac{8}{5}$

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.

$\lim_{n \to 4}\frac{n^2-16}{n^2-3n-4}=\lim_{n \to 4}\frac{(n-4)(n+4)}{(n-4)(n+1)}=\lim_{n \to 4}\frac{n+4}{n+1}=\frac{4+4}{4+1}=\frac{8}{5}$

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Example Question #263 : Limits

Evaluate the limit:

$\lim_{n \to 5}\frac{n^2-4n-5}{n^2-15n+50}$

Possible Answers:

$-\frac{6}{5}$

$-\infty$

Does not exist

$\infty$

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=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.

$\lim_{n \to 5}\frac{n^2-4n-5}{n^2-15n+50}=\lim_{n \to 5}\frac{(n-5)(n+1)}{(n-5)(n-10)}=\lim_{n \to 5}\frac{n+1}{n-10}=\frac{5+1}{5-10}=-\frac{6}{5}$

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Example Question #264 : Limits

Evaluate the limit:

$\lim_{n \to 6}\frac{n^2+2n}{n+6}$

Possible Answers:

Does not exist

$\infty$

$-\infty$

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 not equal zero when n=6 ; so we proceed to insert the value of into the entire equation.

$\lim_{n \to 6}\frac{n^2+2n}{n+6}=\frac{6^2+2\cdot6}{6+6}=\frac{36+12}{12}=\frac{48}{12}=4$

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Example Question #265 : Limits

Evaluate the limit:

$\lim_{n \to 8}2n^2+1$

Possible Answers:

129

Does not exist

$\infty$

$-\infty$

Correct answer:

129

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:

$\lim_{n \to 8}2n^2+1=2\cdot8^2+1=2\cdot64+1=128+1=129$

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Example Question #266 : Limits

Evaluate the limit:

$\lim_{n \to 7}\sqrt{n^2-13}$

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:

$\lim_{n \to 7}\sqrt{n^2-13}=\sqrt{7^2-13}=\sqrt{49-13}=\sqrt{36}=6$

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Example Question #267 : Limits

Evalute the limit:

$\lim_{n \to 9}\sin (0)+\cos (0)-n$

Possible Answers:

$-\infty$

Does not exist

$\infty$

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:

$\lim_{n \to 9}\sin (0)+\cos (0)-n=0+1-9=-8$

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Example Question #268 : Limits

Evaluate the limit:

$\lim_{n \to 10}2n+5n^2-200$

Possible Answers:

320

$\infty$

Does not exist

$-\infty$

Correct answer:

320

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:

$\lim_{n \to 10}2n+5n^2-200=2(10)+5(10)^2-200=20+5\cdot100-200=20+500-200=20+300=320$

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Example Question #269 : Limits

Evaluate the limit:

$\lim_{n \to 11}\frac{n^2-121}{n^2+2n+1}$

Possible Answers:

$-\infty$

$\infty$

Does not exist

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 . In this question, the denominator will not equal zero when n=11 ; so we proceed to insert the value of into the entire equation.

$\lim_{n \to 11}\frac{n^2-121}{n^2+2n+1}=\frac{(11)^2-121}{11^2+2*11+1}=\frac{121-121}{121+22+1}=\frac{0}{144}$

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Example Question #270 : Limits

Evaluate the limit:

$\lim_{n \to 0}\frac{6n^3+4n^2}{5n^4-6n^2}$

Possible Answers:

Does not exist

$-\frac{2}{3}$

$\infty$

$-\infty$

Correct answer:

$-\frac{2}{3}$

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=0 ; 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.

$\lim_{n \to 0}\frac{6n^3+4n^2}{5n^4-6n^2}=\lim_{n \to 0}\frac{2n^2(3n+2)}{n^2(5n^2-6)}=\lim_{n \to 0}\frac{2(3n+2)}{5n^2-6}=\frac{2(3\cdot 0+2)}{5(0)^2-6}=\frac{2(0+2)}{0-6}=\frac{4}{-6}=-\frac{2}{3}$

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Calculus 2 : Limits

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #261 : Limits

Example Question #262 : Limits

Example Question #263 : Limits

Example Question #264 : Limits

Example Question #265 : Limits

Example Question #266 : Limits

Example Question #267 : Limits

Example Question #268 : Limits

Example Question #269 : Limits

Example Question #270 : Limits

All Calculus 2 Resources

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