009C Sample Midterm 2, Problem 1 - Revision history

Kayla Murray at 19:32, 12 November 2017

2017-11-12T19:32:56Z

Kayla Murray: Replaced content with "Evaluate: (a) $\lim _{n\rightarrow \infty} \frac{1}{\big(\frac{n-4}{n}\big)^n}$ (b)
2017-11-05T01:10:33Z

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 01:10, 5 November 2017
Line 4: Line 4:

(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>

+
<hr>

+
(insert picture of handwritten solution)

+
[[009C Sample Midterm 2, Problem 1 Detailed Solution|'''Detailed Solution for this Problem''']]

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

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!Foundations:  

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|-

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|'''1.''' '''L'Hôpital's Rule'''

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|-

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|

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        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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|-

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|

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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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|-

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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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        <math>\begin{array}{rcl}

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\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}}\\

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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''']]

Kayla Murray at 17:31, 18 March 2017

2017-03-18T17:31:31Z

Kayla Murray at 02:02, 27 February 2017

2017-02-27T02:02:53Z

Kayla Murray at 19:06, 26 February 2017

2017-02-26T19:06:24Z

Kayla Murray at 00:18, 19 February 2017

2017-02-19T00:18:44Z

Kayla Murray at 23:56, 15 February 2017

2017-02-15T23:56:22Z

Kayla Murray at 23:54, 15 February 2017

2017-02-15T23:54:57Z

Kayla Murray at 23:51, 15 February 2017

2017-02-15T23:51:03Z

Kayla Murray at 18:57, 13 February 2017

2017-02-13T18:57:42Z

@@ Line 5: / Line 5: @@
 <span class="exam">(b) &nbsp;<math>\sum_{n=1}^\infty \frac{1}{2} \bigg(\frac{1}{4}\bigg)^{n-1} </math>
 <hr>
-(insert picture of handwritten solution)
+[[009C Sample Midterm 2, Problem 1 Solution|'''<u>Solution</u>''']]
 [[009C_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']]

@@ Line 1: / Line 1: @@
 <span class="exam">Evaluate:
-::<span class="exam">a) <math>\lim _{n\rightarrow \infty} \frac{1}{\big(\frac{n-4}{n}\big)^n}</math>
+<span class="exam">(a) <math>\lim _{n\rightarrow \infty} \frac{1}{\big(\frac{n-4}{n}\big)^n}</math>
-::<span class="exam">b) <math>\sum_{n=1}^\infty \frac{1}{2} \bigg(\frac{1}{4}\bigg)^{n-1} </math>

← Older revision		Revision as of 01:10, 5 November 2017
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	<span class="exam">(b)  <math>\sum_{n=1}^\infty \frac{1}{2} \bigg(\frac{1}{4}\bigg)^{n-1} </math>		<span class="exam">(b)  <math>\sum_{n=1}^\infty \frac{1}{2} \bigg(\frac{1}{4}\bigg)^{n-1} </math>
		+	<hr>
		+	(insert picture of handwritten solution)

		+	[[009C Sample Midterm 2, Problem 1 Detailed Solution\|'''<u>Detailed Solution for this Problem</u>''']]

−	~~{\| 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>
−	\|-
−	\|
−	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>
−	\|-
−	\|
−	~~        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>~~
−	\|-
−	~~\|'''2.''' The sum of a convergent geometric series is   <math>\frac{a}{1-r}</math>~~
−	\|-
−	~~\|        where  <math style="vertical-align: 0px">r</math>  is the ratio of the geometric series~~
−	\|-
−	~~\|        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>~~
−	\|-
−	~~\|Now, this limit has the form  <math style="vertical-align: -13px">\frac{0}{0}.</math>~~
−	\|-
−	~~\|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(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:  ~~
−	\|-
−	~~\|Since  <math>\ln y= -4,</math>  we know~~
−	\|-
−	~~\|        <math>y=e^{-4}.</math>~~
−	\|-
−	~~\|Now, we have~~
−	\|-
−	\|
−	~~        <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>~~
−	\|-
−	~~\|Since  <math style="vertical-align: -14px">\|r\|=\frac{1}{4}<1,</math>~~
−	\|-
−	~~\|this series converges.~~
−	\|}
−
−	~~{\| class="mw-collapsible mw-collapsed" style = "text-align:left;"~~
−	~~!Step 2:  ~~
−	\|-
−	~~\|Now, we need to find the sum of this series.~~
−	\|-
−	~~\|The first term of the series is  <math style="vertical-align: -13px">a_1=\frac{1}{2}.</math>~~
−	\|-
−	~~\|Hence, the sum of the series is~~
−	\|-
−	\|
−	~~        <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:  ~~
−	\|-
−	~~\|    '''(a)'''     <math style="vertical-align: -1px">e^{4}</math>~~
−	\|-
−	~~\|    '''(b)'''     <math style="vertical-align: -20px">\frac{2}{3}</math>~~
−	\|}
	[[009C_Sample_Midterm_2\|'''<u>Return to Sample Exam</u>''']]		[[009C_Sample_Midterm_2\|'''<u>Return to Sample Exam</u>''']]

@@ Line 89: / Line 89: @@
 !Step 4: &nbsp;
 |-
-|Since &nbsp;<math>\ln y= -4,</math> we know
+|Since &nbsp;<math>\ln y= -4,</math>&nbsp; we know
 |-
 |&nbsp; &nbsp; &nbsp; &nbsp; <math>y=e^{-4}.</math>

@@ Line 1: / Line 1: @@
 <span class="exam">Evaluate:
-<span class="exam">(a) <math>\lim _{n\rightarrow \infty} \frac{1}{\big(\frac{n-4}{n}\big)^n}</math>
+<span class="exam">(a) &nbsp;<math>\lim _{n\rightarrow \infty} \frac{1}{\big(\frac{n-4}{n}\big)^n}</math>
-<span class="exam">(b) <math>\sum_{n=1}^\infty \frac{1}{2} \bigg(\frac{1}{4}\bigg)^{n-1} </math>
+<span class="exam">(b) &nbsp;<math>\sum_{n=1}^\infty \frac{1}{2} \bigg(\frac{1}{4}\bigg)^{n-1} </math>
@@ Line 12: / Line 12: @@
 |-
+|
-&nbsp; &nbsp; &nbsp; &nbsp; Suppose that <math style="vertical-align: -11px">\lim_{x\rightarrow \infty} f(x)</math>&thinsp; and <math style="vertical-align: -11px">\lim_{x\rightarrow \infty} g(x)</math>&thinsp; are both zero or both <math style="vertical-align: -1px">\pm \infty .</math>
+&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; If <math style="vertical-align: -19px">\lim_{x\rightarrow \infty} \frac{f'(x)}{g'(x)}</math>&thinsp; is finite or&thinsp; <math style="vertical-align: -4px">\pm \infty ,</math>
+&nbsp; &nbsp; &nbsp; &nbsp; If &nbsp;<math style="vertical-align: -19px">\lim_{x\rightarrow \infty} \frac{f'(x)}{g'(x)}</math>&nbsp; is finite or &nbsp;<math style="vertical-align: -4px">\pm \infty ,</math>
 |-
+|
-&nbsp; &nbsp; &nbsp; &nbsp; 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>
+&nbsp; &nbsp; &nbsp; &nbsp; then &nbsp;<math style="vertical-align: -19px">\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>
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp; where <math style="vertical-align: 0px">r</math> is the ratio of the geometric series
+|&nbsp; &nbsp; &nbsp; &nbsp; where &nbsp;<math style="vertical-align: 0px">r</math>&nbsp; is the ratio of the geometric series
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp; and <math style="vertical-align: 0px">a</math> is the first term of the series.
+|&nbsp; &nbsp; &nbsp; &nbsp; and &nbsp;<math style="vertical-align: 0px">a</math>&nbsp; is the first term of the series.
 |}
@@ Line 62: / Line 62: @@
 \end{array}</math>
 |-
-|Now, this limit has the form <math style="vertical-align: -13px">\frac{0}{0}.</math>
+|Now, this limit has the form &nbsp;<math style="vertical-align: -13px">\frac{0}{0}.</math>
 |-
 |Hence, we can use L'Hopital's Rule to calculate this limit.
@@ Line 89: / Line 89: @@
 !Step 4: &nbsp;
 |-
-|Since <math>\ln y= -4,</math> we know
+|Since &nbsp;<math>\ln y= -4,</math> we know
 |-
 |&nbsp; &nbsp; &nbsp; &nbsp; <math>y=e^{-4}.</math>
@@ Line 109: / Line 109: @@
 !Step 1: &nbsp;
 |-
-|First, we not that this is a geometric series with <math style="vertical-align: -14px">r=\frac{1}{4}.</math>
+|First, we not that this is a geometric series with &nbsp;<math style="vertical-align: -14px">r=\frac{1}{4}.</math>
 |-
-|Since <math style="vertical-align: -14px">|r|=\frac{1}{4}<1,</math>
+|Since &nbsp;<math style="vertical-align: -14px">|r|=\frac{1}{4}<1,</math>
 |-
 |this series converges.
@@ Line 121: / Line 121: @@
 |Now, we need to find the sum of this series.
 |-
-|The first term of the series is <math style="vertical-align: -13px">a_1=\frac{1}{2}.</math>
+|The first term of the series is &nbsp;<math style="vertical-align: -13px">a_1=\frac{1}{2}.</math>
 |-
 |Hence, the sum of the series is

@@ Line 26: / Line 26: @@
 |&nbsp; &nbsp; &nbsp; &nbsp; and <math style="vertical-align: 0px">a</math> is the first term of the series.
 |}
 '''Solution:'''
@@ Line 133: / Line 134: @@
 \end{array}</math>
 |}
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"

@@ Line 8: / Line 8: @@
 !Foundations: &nbsp;
 |-
-|L'Hopital's Rule
+|'''1.''' '''L'Hôpital's Rule'''
 |-
-|Sum formula for geometric series
 |-
+|
 |}

@@ Line 49: / Line 49: @@
 \end{array}</math>
 |-
-|Now, this limit has the form <math>\frac{0}{0}.</math>
+|Now, this limit has the form <math style="vertical-align: -13px">\frac{0}{0}.</math>
 |-
 |Hence, we can use L'Hopital's Rule to calculate this limit.
@@ Line 76: / Line 76: @@
 !Step 4: &nbsp;
 |-
-|Since <math>\ln y= -4,</math> <math>y=e^{-4}.</math>
+|Since <math>\ln y= -4,</math> we know
 |-
 |Now, we have
@@ Line 86: / Line 88: @@
 & = & \displaystyle{\frac{1}{e^{-4}}}\\
 &&\\
-& = & \displaystyle{e^4}
+& = & \displaystyle{e^4.}
 \end{array}</math>
 |}
@@ Line 94: / Line 96: @@
 !Step 1: &nbsp;
 |-
-|First, we not that this is a geometric series with <math>r=\frac{1}{4}.</math>
+|First, we not that this is a geometric series with <math style="vertical-align: -14px">r=\frac{1}{4}.</math>
 |-
-|Since <math>|r|=\frac{1}{4}<1,</math>
+|Since <math style="vertical-align: -14px">|r|=\frac{1}{4}<1,</math>
 |-
 |this series converges.
@@ Line 106: / Line 108: @@
 |Now, we need to find the sum of this series.
 |-
-|The first term of the series is <math>a_1=\frac{1}{2}.</math>
+|The first term of the series is <math style="vertical-align: -13px">a_1=\frac{1}{2}.</math>
 |-
 |Hence, the sum of the series is
@@ Line 116: / Line 118: @@
 & = & \displaystyle{\frac{\big(\frac{1}{2}\big)}{\big(\frac{3}{4}\big)}}\\
 &&\\
-& = & \displaystyle{\frac{2}{3}}
+& = & \displaystyle{\frac{2}{3}.}
 \end{array}</math>
 |}
@@ Line 123: / Line 125: @@
 !Final Answer: &nbsp;
 |-
-|&nbsp; &nbsp; '''(a)''' &nbsp; &nbsp; <math>e^{4}</math>
+|&nbsp; &nbsp; '''(a)''' &nbsp; &nbsp; <math style="vertical-align: -1px">e^{4}</math>
 |-
-|&nbsp; &nbsp; '''(b)''' &nbsp; &nbsp; <math>\frac{2}{3}</math>
+|&nbsp; &nbsp; '''(b)''' &nbsp; &nbsp; <math style="vertical-align: -20px">\frac{2}{3}</math>
 |}
 [[009C_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']]