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Calculus 2 : Finding Limits and One-Sided Limits

Study concepts, example questions & explanations for Calculus 2

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Example Questions

Example Question #264 : Limits

Evaluate the limit:

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

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 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:

$-\infty$

Does not exist

$\infty$

129

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:

$\infty$

$-\infty$

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 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:

$-\infty$

$\infty$

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 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:

$\infty$

Does not exist

320

$-\infty$

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:

$-\frac{2}{3}$

Does not exist

$-\infty$

$\infty$

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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Example Question #271 : Limits

Considering the following piecewise function, what is $\lim_{x \rightarrow 2^+} y,$ $y= \begin{Bmatrix}3 & x < 2 \\ -2 & x \geq 2\end{Bmatrix}$

Possible Answers:

Does not exist

Correct answer:

Explanation:

In general, when you are looking for $\lim_{x \rightarrow a^+} y$ you are looking to see whether the limit of y exists to the right, and if it does, what is the value.

Solution:

In this case, we want to see the limit at x=2 , from the right. The limit exists, and the value is

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Example Question #272 : Limits

Considering the following piecewise function, what is $\lim_{x \rightarrow 6^-} y$ ,

$y = \begin{Bmatrix}x+2 & x < 6 \\ x^2 & x \geq 6\end{Bmatrix}$

Possible Answers:

Does not exist

Correct answer:

Explanation:

In general, when you are looking fo r $\lim_{x \rightarrow a^-} y$ you are looking to see whether the limit of y exists to the left, and if it does, what is the value.

Solution:

In this case, we want to see the limit at x=6 , from the left. The limit exists, as reflected by the function x+2

6+2 = 8

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Example Question #273 : Limits

Considering the following piecewise function, what is $\lim_{x \rightarrow 5^-} y$ , $y = \begin{Bmatrix}x+2 & x < 5 \\ x^2 + 4 & x = 5 \\x^2 + x & x \geq 5\end{Bmatrix}$

Possible Answers:

Does not exist

Correct answer:

Explanation:

In general, when you are looking for $\lim_{x \rightarrow a^-}$ y you are looking to see whether the limit of y exists to the left, and if it does, what is the value.

Solution:

In this case, we want to see the limit at x=5 , from the left. The limit exists and corresponds to the function x + 2 .

Plug in 5 for x and solve to get the value of the limit from the left.

5+2 = 7

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Calculus 2 : Finding Limits and One-Sided Limits

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

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

Example Question #271 : Limits

Example Question #272 : Limits

Example Question #273 : Limits

All Calculus 2 Resources

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