009C Sample Final 1, Problem 1 - Revision history

MathAdmin: Created page with "Compute ::a) $\lim_{n\rightarrow \infty} \frac{3-2n^2}{5n^2+n+1}$ ::b)
2016-04-19T01:05:35Z

Created page with "<span class="exam">Compute ::<span class="exam">a) <math style="vertical-align: -12px">\lim_{n\rightarrow \infty} \frac{3-2n^2}{5n^2+n+1}</math> ::<span class="exam">b) <mat..."

New page
<span class="exam">Compute

::<span class="exam">a) <math style="vertical-align: -12px">\lim_{n\rightarrow \infty} \frac{3-2n^2}{5n^2+n+1}</math>

::<span class="exam">b) <math style="vertical-align: -12px">\lim_{n\rightarrow \infty} \frac{\ln n}{\ln 3n}</math>

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
|-
|Recall:
|-
|
::'''L'Hopital's Rule'''
|-
|
::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>
|-
|
::If <math>\lim_{x\rightarrow \infty} \frac{f'(x)}{g'(x)}</math> is finite or <math style="vertical-align: -1px">\pm \infty ,</math>
|-
|
::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 switch to the limit to <math style="vertical-align: 0px">x</math> so that we can use L'Hopital's rule.
|-
|So, we have
|-
|
::<math>\begin{array}{rcl}
\displaystyle{\lim_{x \rightarrow \infty}\frac{3-2x^2}{5x^2 + x +1}} & \overset{l'H}{=} & \displaystyle{\lim_{x \rightarrow \infty}\frac{-4x}{10x+1}}\\
&&\\
& \overset{l'H}{=} & \displaystyle{\frac{-4}{10}}\\
&&\\
& = & \displaystyle{\frac{-2}{5}}.
\end{array}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Hence, we have
|-
|
::<math>\lim_{n\rightarrow \infty} \frac{3-2n^2}{5n^2+n+1}=\frac{-2}{5}.</math>
|}

'''(b)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|Again, we switch to the limit to <math style="vertical-align: 0px">x</math> so that we can use L'Hopital's rule.
|-
|So, we have
|-
|
::<math>\begin{array}{rcl}
\displaystyle{\lim_{x \rightarrow \infty}\frac{\ln x}{\ln(3x)}} & \overset{l'H}{=} & \displaystyle{\lim_{x \rightarrow \infty}\frac{\frac{1}{x}}{\frac{3}{3x}}}\\
&&\\
& = & \displaystyle{\lim_{x \rightarrow \infty} 1}\\
&&\\
& = & 1.
\end{array}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Hence, we have
|-
|
::<math>\lim_{n\rightarrow \infty} \frac{\ln n}{\ln 3n}=1.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|   '''(a)''' <math style="vertical-align: -14px">\frac{-2}{5}</math>
|-
|   '''(b)''' <math style="vertical-align: -3px">1</math>
|}
[[009C_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]