009C Sample Midterm 2, Problem 1 - Revision history

MathAdmin: Replaced content with "Evaluate: (a) $\lim _{n\rightarrow \infty} \frac{1}{\big(\frac{n-4}{n}\big)^n}$ (b) 2017-11-12T19:35:45Z Replaced content with "Evaluate: (a) <math>\lim _{n\rightarrow \infty} \frac{1}{\big(\frac{n-4}{n}\big)^n}</math> (b) <math..."

← Older revision Revision as of 19:35, 12 November 2017
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(b)  <math>\sum_{n=1}^\infty \frac{1}{2} \bigg(\frac{1}{4}\bigg)^{n-1} </math>

(b)  <math>\sum_{n=1}^\infty \frac{1}{2} \bigg(\frac{1}{4}\bigg)^{n-1} </math>

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<hr>

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[[009C Sample Midterm 2, Problem 1 Solution|'''Solution''']]

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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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[[009C Sample Midterm 2, Problem 1 Detailed Solution|'''Detailed Solution''']]

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!Foundations:  

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|-

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|'''1.''' '''L'Hôpital's Rule, Part 2'''

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|-

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|

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        Let  <math style="vertical-align: -5px">f</math>  and  <math style="vertical-align: -5px">g</math>  be differentiable functions on the open interval  <math style="vertical-align: -5px">(a,\infty)</math>  for some value  <math style="vertical-align: -4px">a,</math> 

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|-

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|        where  <math style="vertical-align: -5px">g'(x)\ne 0</math>  on  <math style="vertical-align: -5px">(a,\infty)</math>  and  <math style="vertical-align: -18px">\lim_{x\rightarrow \infty} \frac{f(x)}{g(x)}</math>  returns either  <math style="vertical-align: -15px">\frac{0}{0}</math>  or  <math style="vertical-align: -15px">\frac{\infty}{\infty}.</math> 

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|-

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|       Then,   <math style="vertical-align: -18px">\lim_{x\rightarrow \infty} \frac{f(x)}{g(x)}=\lim_{x\rightarrow \infty} \frac{f'(x)}{g'(x)}.</math>

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|-

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|'''2.''' The sum of a convergent geometric series is   <math>\frac{a}{1-r}</math>

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|-

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|        where  <math style="vertical-align: 0px">r</math>  is the ratio of the geometric series

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|-

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|        and  <math style="vertical-align: 0px">a</math>  is the first term of the series.

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|}

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'''Solution:'''

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'''(a)'''

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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"

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!Step 1:  

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|-

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|Let

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        <math>\begin{array}{rcl}

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\displaystyle{y} & = & \displaystyle{\lim_{n\rightarrow \infty} \bigg(\frac{n-4}{n}\bigg)^n}\\

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&&\\

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& = & \displaystyle{\lim_{n\rightarrow \infty} \bigg(1-\frac{4}{n}\bigg)^n.}

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\end{array}</math>

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|-

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|We then take the natural log of both sides to get

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|-

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|        <math>\ln y = \ln\bigg(\lim_{n\rightarrow \infty} \bigg(1-\frac{4}{n}\bigg)^n\bigg).</math>

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|}

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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"

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!Step 2:  

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|-

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|We can interchange limits and continuous functions.

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|-

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|Therefore, we have

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|-

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|

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        <math>\begin{array}{rcl}

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\displaystyle{\ln y} & = & \displaystyle{\lim_{n\rightarrow \infty} \ln \bigg(1-\frac{4}{n}\bigg)^n}\\

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&&\\

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& = & \displaystyle{\lim_{n\rightarrow \infty} n\ln\bigg(1-\frac{4}{n}\bigg)}\\

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&&\\

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& = & \displaystyle{\lim_{n\rightarrow \infty} \frac{\ln \bigg(1-\frac{4}{n}\bigg)}{\frac{1}{n}}.}

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\end{array}</math>

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|-

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|Now, this limit has the form  <math style="vertical-align: -13px">\frac{0}{0}.</math>

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|-

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|Hence, we can use L'Hopital's Rule to calculate this limit.

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|}

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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"

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!Step 3:  

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|-

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|Now, we have

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|-

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|

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        <math>\begin{array}{rcl}

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\displaystyle{\ln y } & = & \displaystyle{\lim_{n\rightarrow \infty} \frac{\ln \bigg(1-\frac{4}{n}\bigg)}{\frac{1}{n}}}\\

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&&\\

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& = & \displaystyle{\lim_{x\rightarrow \infty} \frac{\ln \bigg(1-\frac{4}{x}\bigg)}{\frac{1}{x}}}\\

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&&\\

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& \overset{L'H}{=} & \displaystyle{\lim_{x\rightarrow \infty} \frac{\frac{1}{\big(1-\frac{4}{x}\big)}\frac{4}{x^2}}{\big(-\frac{1}{x^2}\big)}}\\

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&&\\

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& = & \displaystyle{\lim_{x\rightarrow \infty} \frac{-4x}{x-4}}\\

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&&\\

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& = & \displaystyle{-4.}

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\end{array}</math>

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|}

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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"

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!Step 4:  

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|-

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|Since  <math>\ln y= -4,</math>  we know

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|-

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|        <math>y=e^{-4}.</math>

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|-

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|Now, we have

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|-

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|

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        <math>\begin{array}{rcl}

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\displaystyle{\lim _{n\rightarrow \infty} \frac{1}{\big(\frac{n-4}{n}\big)^n}} & = & \displaystyle{\frac{\displaystyle{\lim_{n\rightarrow \infty} 1}}{\displaystyle{\lim_{n\rightarrow \infty} \bigg(\frac{n-4}{n}\bigg)^n}}}\\

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&&\\

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& = & \displaystyle{\frac{1}{e^{-4}}}\\

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&&\\

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& = & \displaystyle{e^4.}

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\end{array}</math>

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|}

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'''(b)'''

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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"

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!Step 1:  

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|-

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|First, we not that this is a geometric series with  <math style="vertical-align: -14px">r=\frac{1}{4}.</math>

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|-

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|Since  <math style="vertical-align: -14px">|r|=\frac{1}{4}<1,</math>

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|-

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|this series converges.

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|}

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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"

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!Step 2:  

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|-

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|Now, we need to find the sum of this series.

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|-

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|The first term of the series is  <math style="vertical-align: -13px">a_1=\frac{1}{2}.</math>

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|-

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|Hence, the sum of the series is

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|-

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|

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        <math>\begin{array}{rcl}

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\displaystyle{\frac{a_1}{1-r}} & = & \displaystyle{\frac{\frac{1}{2}}{1-\frac{1}{4}}}\\

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&&\\

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& = & \displaystyle{\frac{\big(\frac{1}{2}\big)}{\big(\frac{3}{4}\big)}}\\

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&&\\

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& = & \displaystyle{\frac{2}{3}.}

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\end{array}</math>

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|}

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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"

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!Final Answer:  

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|-

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|    '''(a)'''     <math style="vertical-align: -1px">e^{4}</math>

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|-

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|    '''(b)'''     <math style="vertical-align: -20px">\frac{2}{3}</math>

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|}

[[009C_Sample_Midterm_2|'''Return to Sample Exam''']]

[[009C_Sample_Midterm_2|'''Return to Sample Exam''']]

MathAdmin at 16:53, 16 April 2017

2017-04-16T16:53:37Z

MathAdmin at 17:30, 14 April 2017

2017-04-14T17:30:17Z

MathAdmin: Created page with "Evaluate: (a) $\lim _{n\rightarrow \infty} \frac{1}{\big(\frac{n-4}{n}\big)^n}$ (b) $\su..."$

2017-04-09T19:29:42Z

Created page with "Evaluate: (a) <math>\lim _{n\rightarrow \infty} \frac{1}{\big(\frac{n-4}{n}\big)^n}</math> (b) <math>\su..."

New page

Evaluate:

(a)  <math>\lim _{n\rightarrow \infty} \frac{1}{\big(\frac{n-4}{n}\big)^n}</math>

(b)  <math>\sum_{n=1}^\infty \frac{1}{2} \bigg(\frac{1}{4}\bigg)^{n-1} </math>

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
|-
|'''1.''' '''L'Hôpital's Rule'''
|-
|
        Suppose that  <math style="vertical-align: -11px">\lim_{x\rightarrow \infty} f(x)</math>  and  <math style="vertical-align: -11px">\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 style="vertical-align: -19px">\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 style="vertical-align: -19px">\lim_{x\rightarrow \infty} \frac{f(x)}{g(x)}\,=\,\lim_{x\rightarrow \infty} \frac{f'(x)}{g'(x)}.</math>
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|'''2.''' The sum of a convergent geometric series is   <math>\frac{a}{1-r}</math>
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|        where  <math style="vertical-align: 0px">r</math>  is the ratio of the geometric series
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|        and  <math style="vertical-align: 0px">a</math>  is the first term of the series.
|}

'''Solution:'''

'''(a)'''
{| 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-4}{n}\bigg)^n}\\
&&\\
& = & \displaystyle{\lim_{n\rightarrow \infty} \bigg(1-\frac{4}{n}\bigg)^n.}
\end{array}</math>
|-
|We then take the natural log of both sides to get
|-
|        <math>\ln y = \ln\bigg(\lim_{n\rightarrow \infty} \bigg(1-\frac{4}{n}\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(1-\frac{4}{n}\bigg)^n}\\
&&\\
& = & \displaystyle{\lim_{n\rightarrow \infty} n\ln\bigg(1-\frac{4}{n}\bigg)}\\
&&\\
& = & \displaystyle{\lim_{n\rightarrow \infty} \frac{\ln \bigg(1-\frac{4}{n}\bigg)}{\frac{1}{n}}.}
\end{array}</math>
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|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
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|
        <math>\begin{array}{rcl}
\displaystyle{\ln y } & = & \displaystyle{\lim_{n\rightarrow \infty} \frac{\ln \bigg(1-\frac{4}{n}\bigg)}{\frac{1}{n}}}\\
&&\\
& = & \displaystyle{\lim_{x\rightarrow \infty} \frac{\ln \bigg(1-\frac{4}{x}\bigg)}{\frac{1}{x}}}\\
&&\\
& \overset{L'H}{=} & \displaystyle{\lim_{x\rightarrow \infty} \frac{\frac{1}{\big(1-\frac{4}{x}\big)}\frac{4}{x^2}}{\big(-\frac{1}{x^2}\big)}}\\
&&\\
& = & \displaystyle{\lim_{x\rightarrow \infty} \frac{-4x}{x-4}}\\
&&\\
& = & \displaystyle{-4.}
\end{array}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 4:  
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|Since  <math>\ln y= -4,</math>  we know
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|        <math>y=e^{-4}.</math>
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|Now, we have
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|
        <math>\begin{array}{rcl}
\displaystyle{\lim _{n\rightarrow \infty} \frac{1}{\big(\frac{n-4}{n}\big)^n}} & = & \displaystyle{\frac{\lim_{n\rightarrow \infty} 1}{\lim_{n\rightarrow \infty} \big(\frac{n-4}{n}\big)^n}}\\
&&\\
& = & \displaystyle{\frac{1}{e^{-4}}}\\
&&\\
& = & \displaystyle{e^4.}
\end{array}</math>
|}

'''(b)'''
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|First, we not that this is a geometric series with  <math style="vertical-align: -14px">r=\frac{1}{4}.</math>
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|Since  <math style="vertical-align: -14px">|r|=\frac{1}{4}<1,</math>
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|this series converges.
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Now, we need to find the sum of this series.
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|The first term of the series is  <math style="vertical-align: -13px">a_1=\frac{1}{2}.</math>
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|Hence, the sum of the series is
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|
        <math>\begin{array}{rcl}
\displaystyle{\frac{a_1}{1-r}} & = & \displaystyle{\frac{\frac{1}{2}}{1-\frac{1}{4}}}\\
&&\\
& = & \displaystyle{\frac{\big(\frac{1}{2}\big)}{\big(\frac{3}{4}\big)}}\\
&&\\
& = & \displaystyle{\frac{2}{3}.}
\end{array}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
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|    '''(a)'''     <math style="vertical-align: -1px">e^{4}</math>
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|    '''(b)'''     <math style="vertical-align: -20px">\frac{2}{3}</math>
|}
[[009C_Sample_Midterm_2|'''Return to Sample Exam''']]

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 <span class="exam">(b) &nbsp;<math>\sum_{n=1}^\infty \frac{1}{2} \bigg(\frac{1}{4}\bigg)^{n-1} </math>
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
+[[009C Sample Midterm 2, Problem 1 Detailed Solution|'''<u>Detailed Solution</u>''']]
 [[009C_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']]

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 !Foundations: &nbsp;
 |-
-|'''1.''' '''L'Hôpital's Rule'''
+|'''1.''' '''L'Hôpital's Rule, Part 2'''
 |-
+|
-&nbsp; &nbsp; &nbsp; &nbsp; Suppose that &nbsp;<math style="vertical-align: -11px">\lim_{x\rightarrow \infty} f(x)</math>&nbsp; and &nbsp;<math style="vertical-align: -11px">\lim_{x\rightarrow \infty} g(x)</math>&nbsp; are both zero or both &nbsp;<math style="vertical-align: -1px">\pm \infty .</math>
+&nbsp; &nbsp; &nbsp; &nbsp; Let &nbsp;<math style="vertical-align: -5px">f</math>&nbsp; and &nbsp;<math style="vertical-align: -5px">g</math>&nbsp; be differentiable functions on the open interval &nbsp;<math style="vertical-align: -5px">(a,\infty)</math>&nbsp; for some value &nbsp;<math style="vertical-align: -4px">a,</math>&nbsp;
 |-
+|&nbsp; &nbsp; &nbsp; &nbsp; where &nbsp;<math style="vertical-align: -5px">g'(x)\ne 0</math>&nbsp; on &nbsp;<math style="vertical-align: -5px">(a,\infty)</math>&nbsp; and &nbsp;<math style="vertical-align: -18px">\lim_{x\rightarrow \infty} \frac{f(x)}{g(x)}</math>&nbsp; returns either &nbsp;<math style="vertical-align: -15px">\frac{0}{0}</math>&nbsp; or &nbsp;<math style="vertical-align: -15px">\frac{\infty}{\infty}.</math>&nbsp;
 |-
+|&nbsp; &nbsp; &nbsp; &nbsp;Then, &nbsp; <math style="vertical-align: -18px">\lim_{x\rightarrow \infty} \frac{f(x)}{g(x)}=\lim_{x\rightarrow \infty} \frac{f'(x)}{g'(x)}.</math>
 |-
 |'''2.''' The sum of a convergent geometric series is &nbsp; <math>\frac{a}{1-r}</math>

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+|
 &nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
-\displaystyle{\lim _{n\rightarrow \infty} \frac{1}{\big(\frac{n-4}{n}\big)^n}} & = & \displaystyle{\frac{\lim_{n\rightarrow \infty} 1}{\lim_{n\rightarrow \infty} \big(\frac{n-4}{n}\big)^n}}\\
+\displaystyle{\lim _{n\rightarrow \infty} \frac{1}{\big(\frac{n-4}{n}\big)^n}} & = & \displaystyle{\frac{\displaystyle{\lim_{n\rightarrow \infty} 1}}{\displaystyle{\lim_{n\rightarrow \infty} \bigg(\frac{n-4}{n}\bigg)^n}}}\\
 &&\\
 & = & \displaystyle{\frac{1}{e^{-4}}}\\

009C Sample Midterm 2, Problem 1 - Revision history

MathAdmin at 16:53, 16 April 2017

MathAdmin at 17:30, 14 April 2017

MathAdmin: Created page with "Evaluate: (a) \lim _{n\rightarrow \infty} \frac{1}{\big(\frac{n-4}{n}\big)^n} (b) \su..."

MathAdmin: Created page with "Evaluate: (a) $\lim _{n\rightarrow \infty} \frac{1}{\big(\frac{n-4}{n}\big)^n}$ (b) $\su..."$