009C Sample Final 2, Problem 1 - Revision history

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Created page with "<span class="exam">Test if the following sequences converge or diverge. Also find the limit of each convergent sequence. <span class="exam">(a) <math style="vertical-al..."

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<span class="exam">Test if the following sequences converge or diverge. Also find the limit of each convergent sequence.

<span class="exam">(a)  <math style="vertical-align: -16px">a_n=\frac{\ln(n)}{\ln(n+1)}</math>

<span class="exam">(b)  <math style="vertical-align: -15px">a_n=\bigg(\frac{n}{n+1}\bigg)^n</math>

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
|-
|'''L'Hopital's Rule'''
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|
        Suppose that  <math>\lim_{x\rightarrow \infty} f(x)</math>   and  <math>\lim_{x\rightarrow \infty} g(x)</math>   are both zero or both   <math style="vertical-align: -1px">\pm \infty .</math>
|-
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       If  <math>\lim_{x\rightarrow \infty} \frac{f'(x)}{g'(x)}</math>   is finite or   <math style="vertical-align: -4px">\pm \infty ,</math>
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       then  <math>\lim_{x\rightarrow \infty} \frac{f(x)}{g(x)}=\lim_{x\rightarrow \infty} \frac{f'(x)}{g'(x)}.</math>
|}

'''Solution:'''

'''(a)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|First, we notice that  <math>\lim_{n\rightarrow \infty} \frac{\ln(n)}{\ln(n+1)}</math>  has the form  <math>\frac{\infty}{\infty}.</math>
|-
|So, we can use L'Hopital's Rule. To begin, we write
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|       <math>\lim_{n\rightarrow \infty}\frac{\ln(n)}{\ln(n+1)}=\lim_{x\rightarrow \infty} \frac{\ln(x)}{\ln(x+1)}.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Now, using L'Hopital's rule, we get
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|        <math>\begin{array}{rcl}
\displaystyle{\lim_{n\rightarrow \infty}\frac{\ln(n)}{\ln(n+1)}} & \overset{L'H}{=} & \displaystyle{\lim_{x\rightarrow \infty} \frac{\frac{1}{x}}{\frac{1}{x+1}}}\\
&&\\
& = & \displaystyle{\lim_{x\rightarrow \infty} \frac{x+1}{x}}\\
&&\\
& = & \displaystyle{\lim_{x\rightarrow \infty} 1+\frac{1}{x}}\\
&&\\
& = & \displaystyle{1.}
\end{array}</math>
|}

'''(b)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|Let
        <math>\begin{array}{rcl}
\displaystyle{y} & = & \displaystyle{\lim_{n\rightarrow \infty} \bigg(\frac{n}{n+1}\bigg)^n.}
\end{array}</math>
|-
|We then take the natural log of both sides to get
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|        <math>\ln y = \ln\bigg(\lim_{n\rightarrow \infty} \bigg(\frac{n}{n+1}\bigg)^n\bigg).</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|We can interchange limits and continuous functions.
|-
|Therefore, we have
|-
|
        <math>\begin{array}{rcl}
\displaystyle{\ln y} & = & \displaystyle{\lim_{n\rightarrow \infty} \ln \bigg(\frac{n}{n+1}\bigg)^n}\\
&&\\
& = & \displaystyle{\lim_{n\rightarrow \infty} n\ln\bigg(\frac{n}{n+1}\bigg)}\\
&&\\
& = & \displaystyle{\lim_{n\rightarrow \infty} \frac{\ln \bigg(\frac{n}{n+1}\bigg)}{\frac{1}{n}}.}
\end{array}</math>
|-
|Now, this limit has the form  <math style="vertical-align: -13px">\frac{0}{0}.</math>
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|Hence, we can use L'Hopital's Rule to calculate this limit.
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 3:  
|-
|Now, we have
|-
|
        <math>\begin{array}{rcl}
\displaystyle{\ln y } & = & \displaystyle{\lim_{n\rightarrow \infty} \frac{\ln \bigg(\frac{n}{n+1}\bigg)}{\frac{1}{n}}}\\
&&\\
& = & \displaystyle{\lim_{x\rightarrow \infty} \frac{\ln \bigg(\frac{x}{x+1}\bigg)}{\frac{1}{x}}}\\
&&\\
& \overset{L'H}{=} & \displaystyle{\lim_{x\rightarrow \infty} \frac{\frac{x+1}{x}\frac{1}{(x+1)^2}}{\big(-\frac{1}{x^2}\big)}}\\
&&\\
& = & \displaystyle{\lim_{x\rightarrow \infty} \frac{-x^2}{x(x+1)}}\\
&&\\
& = & \displaystyle{-1.}
\end{array}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 4:  
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|Since  <math>\ln y= -1,</math>  we know
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|        <math>y=e^{-1}.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
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|   '''(a)'''     <math>1</math>
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|   '''(b)'''     <math>e^{-1}</math>
|}
[[009C_Sample_Final_2|'''<u>Return to Sample Exam</u>''']]