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Calculus 2 : Limits

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Example Questions

Example Question #251 : Calculus Ii

Evaluate the limit:

$\lim_{n \to -5}\frac{n^2+4n-5}{n^2+7n+10}$

Possible Answers:

$-\infty$

Does Not Exist

$\infty$

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 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.

$\lim_{n \to -5}\frac{n^2+4n-5}{n^2+7n+10}=\lim_{n \to -5}\frac{(n+5)(n-1)}{(n+5)(n+2)} \\ \\ =\lim_{n \to -5}\frac{n-1}{n+2}\\\\=\frac{-5-1}{-5+2}=\frac{-6}{-3}=2$

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Example Question #251 : Calculus Ii

Evaluate the limit:

$\lim_{n \to -5}(5+n)(6-n)$

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:

$\lim_{n \to -5}(5+n)(6-n)=(5+(-5))(6-(-5))=(5-5)(6+5)=0\cdot11=0$

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Example Question #252 : Limits

Evaluate the limit:

$\lim_{n \to 6}\frac{3n^2+15n-18}{6n^2+12n-288}$

Possible Answers:

$\frac{1}{4}$

$\infty$

$-\infty$

Does Not Exist

Correct answer:

$\frac{1}{4}$

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=6 ; 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.

$\lim_{n \to 6}\frac{3n^2+15n-18}{6n^2+12n-288}=\lim_{n \to 6}\frac{3(n^2-5n-6)}{6(n^2+2n-48)} \\ \\ =\lim_{n \to 6}\frac{3(n-6)(n+1)}{6(n-6)(n+8)}=\lim_{n \to 6}\frac{n+1}{2(n+8)} \\ \\ =\frac{6+1}{2(6+8)}=\frac{7}{2(14)}=\frac{1}{2\cdot2}=\frac{1}{4}$

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Example Question #252 : Calculus Ii

Evaluate the limit:

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

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:

$\lim_{n \to 6}n^2+n=(6^2)+6=36+6=42$

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Example Question #255 : Calculus Ii

Evaluate the limit:

$\lim_{n \to -6}\frac{n^2+9n+18}{n^2-36}$

Possible Answers:

$\infty$

Does Not Exist

$-\infty$

$\frac{1}{4}$

Correct answer:

$\frac{1}{4}$

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 0. In this question, the denominator will equal zero when n=-6 ; 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.

$\lim_{n \to -6}\frac{n^2+9n+18}{n^2-36}=\lim_{n \to -6}\frac{(n+6)(n+3)}{(n+6)(n-6)} \\ \\ =\lim_{n \to -6}\frac{n+3}{n-6}=\frac{-6+3}{-6-6} \\ \\ =\frac{-3}{-12}=\frac{1}{4}$

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Example Question #253 : Calculus Ii

Evaluate the limit:

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

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:

$\lim_{n \to -6}3n^2+2n-6=3(-6)^2+2(-6)-6=3\cdot36-12-6=108-18=90$

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Example Question #257 : Calculus Ii

Evaluate the following limit:

$\lim_{x\rightarrow 3^{-} } f(x), f(x)=\left\{\begin{matrix} e^x, x< 3 \\ \ln(x-3) , x\geq 3\end{matrix}\right.$

Possible Answers:

$\infty$

The limit does not exist

$-\infty$

$\frac{1}{e^3}$

e^3

Correct answer:

e^3

Explanation:

In order to evaluate the limit, we must see whether the limit is right or left sided. The negative sign exponent on the indicates we are approaching from the left side, or with values slightly less than . The part of the piecewise function corresponding to these values is the first function, and evaluated at , we get e^3

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Example Question #258 : Calculus Ii

Evaluate the limit:

$\lim_{x\rightarrow \infty }\frac{8x^4+4x^3}{3x^4+x^3+2}$

Possible Answers:

$\frac{8}{3}$

$\infty$

$-\infty$

Correct answer:

$\frac{8}{3}$

Explanation:

To evaluate the limit as approaches infinity, we must first take out a term consisting of the highest power term divided by itself (, essentially:

$\lim_{x\rightarrow \infty }\frac{x^4}{x^4}(\frac{8+4x^{-1}}{3+x^{-1}+2x^{-2}})$

When the term we created becomes , and the terms with negative exponents become zero (as approaches infinity, the fraction with infinity in the denominator equals zero), we are left with our answer, $\frac{8}{3}$

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Example Question #259 : Calculus Ii

Evaluate the limit:

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

Possible Answers:

$\infty$

Does not exist

$-\infty$

$\frac{3}{4}$

Correct answer:

$\frac{3}{4}$

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=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 into the remaining equation.

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

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Example Question #260 : Calculus Ii

Evaluate the limit:

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

Possible Answers:

$-\frac{1}{3}$

Does not exist

$\infty$

$-\infty$

Correct answer:

$-\frac{1}{3}$

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=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 into the remaining equation.

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

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Calculus 2 : Limits

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #251 : Calculus Ii

Example Question #251 : Calculus Ii

Example Question #252 : Limits

Example Question #252 : Calculus Ii

Example Question #255 : Calculus Ii

Example Question #253 : Calculus Ii

Example Question #257 : Calculus Ii

Example Question #258 : Calculus Ii

Example Question #259 : Calculus Ii

Example Question #260 : Calculus Ii

All Calculus 2 Resources

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