Difference between revisions of "009C Sample Midterm 1, Problem 1"

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::<math>a_n=\frac{\ln n}{n}</math>
 
::<math>a_n=\frac{\ln n}{n}</math>
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<hr>
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[[009C Sample Midterm 1, Problem 1 Solution|'''<u>Solution</u>''']]
  
  
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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[[009C Sample Midterm 1, Problem 1 Detailed Solution|'''<u>Detailed Solution</u>''']]
!Foundations: &nbsp;
 
|-
 
|'''L'Hôpital's Rule, Part 2'''
 
|-
 
|
 
&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>
 
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'''Solution:'''
 
 
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!Step 1: &nbsp;
 
|-
 
|First, notice that
 
|-
 
|&nbsp; &nbsp; &nbsp; &nbsp; <math>\lim_{n\rightarrow \infty} \ln n =\infty</math>
 
|-
 
|and
 
|-
 
|&nbsp; &nbsp; &nbsp; &nbsp; <math>\lim_{n\rightarrow \infty} n=\infty.</math>
 
|-
 
|Therefore, the limit has the form &nbsp;<math style="vertical-align: -11px">\frac{\infty}{\infty},</math>
 
|-
 
|which means that we can use L'Hopital's Rule to calculate this limit.
 
|}
 
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
!Step 2: &nbsp;
 
|-
 
|First, switch to the variable &nbsp;<math style="vertical-align: 0px">x</math> &nbsp; so that we have functions and
 
|-
 
|can take derivatives. Thus, using L'Hopital's Rule, we have
 
|-
 
|&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 
\displaystyle{\lim_{n\rightarrow \infty} \frac{\ln n}{n}} & = & \displaystyle{\lim_{x\rightarrow \infty} \frac{\ln x}{x}}\\
 
&&\\
 
& \overset{L'H}{=} & \displaystyle{\lim_{x\rightarrow \infty} \frac{\big(\frac{1}{x}\big)}{1}}\\
 
&&\\
 
& = & \displaystyle{0.}
 
\end{array}</math>
 
|}
 
 
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
!Final Answer: &nbsp;
 
|-
 
|&nbsp; &nbsp; &nbsp; &nbsp; The sequence converges. The limit of the sequence is &nbsp;<math style="vertical-align: 0px">0.</math>
 
|}
 
 
[[009C_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']]
 
[[009C_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']]

Latest revision as of 22:52, 11 November 2017

Does the following sequence converge or diverge?

If the sequence converges, also find the limit of the sequence.

Be sure to jusify your answers!


Solution


Detailed Solution


Return to Sample Exam