Difference between revisions of "009C Sample Final 3, Problem 1"

Latest revision as of 18:39, 2 December 2017

Which of the following sequences Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (a_n)_{n\ge 1}} converges? Which diverges? Give reasons for your answers!

(a) Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a_n=\bigg(1+\frac{1}{2n}\bigg)^n}

(b) Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a_n=\cos(n\pi)\bigg(\frac{1+n}{n}\bigg)^n}

Solution

Detailed Solution

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@@ Line 5: / Line 5: @@
 <span class="exam">(b) &nbsp;<math style="vertical-align: -15px">a_n=\cos(n\pi)\bigg(\frac{1+n}{n}\bigg)^n</math>
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
+<hr>
-!Foundations: &nbsp;
+[[009C Sample Final 3, Problem 1 Solution|'''<u>Solution</u>''']]
-|-
-|'''L'Hôpital's Rule'''
-|-
-|
-&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 &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 &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>
-|}
-'''Solution:'''
+[[009C Sample Final 3, Problem 1 Detailed Solution|'''<u>Detailed Solution</u>''']]
-'''(a)'''
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 1: &nbsp;
-|-
-|Let
-&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
-\displaystyle{y} & = & \displaystyle{\lim_{n\rightarrow \infty} \bigg(1+\frac{1}{2n}\bigg)^n.}
-\end{array}</math>
-|-
-|We then take the natural log of both sides to get
-|-
-|&nbsp; &nbsp; &nbsp; &nbsp; <math>\ln y = \ln\bigg(\lim_{n\rightarrow \infty} \bigg(1+\frac{1}{2n}\bigg)^n\bigg).</math>
-|}
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 2: &nbsp;
-|-
-|We can interchange limits and continuous functions.
-|-
-|Therefore, we have
-|-
-|
-&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
-\displaystyle{\ln y} & = & \displaystyle{\lim_{n\rightarrow \infty} \ln \bigg(1+\frac{1}{2n}\bigg)^n}\\
-&&\\
-& = & \displaystyle{\lim_{n\rightarrow \infty} n\ln\bigg(1+\frac{1}{2n}\bigg)}\\
-&&\\
-& = & \displaystyle{\lim_{n\rightarrow \infty} \frac{\ln \bigg(1+\frac{1}{2n}\bigg)}{\frac{1}{n}}.}
-\end{array}</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.
-|}
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 3: &nbsp;
-|-
-|Now, we have
-|-
-|
-&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
-\displaystyle{\ln y } & = & \displaystyle{\lim_{n\rightarrow \infty} \frac{\ln \bigg(1+\frac{1}{2n}\bigg)}{\frac{1}{n}}}\\
-&&\\
-& = & \displaystyle{\lim_{x\rightarrow \infty} \frac{\ln \bigg(1+\frac{1}{2x}\bigg)}{\frac{1}{x}}}\\
-&&\\
-& \overset{L'H}{=} & \displaystyle{\lim_{x\rightarrow \infty} \frac{\frac{2x}{2x+1}\big(\frac{-1}{2x^2}\big)}{\big(-\frac{1}{x^2}\big)}}\\
-&&\\
-& = & \displaystyle{\lim_{x\rightarrow \infty} \frac{1}{2}\bigg(\frac{2x}{2x+1}\bigg)}\\
-&&\\
-& = & \displaystyle{\frac{1}{2}.}
-\end{array}</math>
-|}
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 4: &nbsp;
-|-
-|Since &nbsp;<math style="vertical-align: -13px">\ln y= \frac{1}{2},</math>&nbsp; we know
-|-
-|&nbsp; &nbsp; &nbsp; &nbsp; <math>y=e^{\frac{1}{2}}.</math>
-|}
-'''(b)'''
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 1: &nbsp;
-|-
-|First, we have
-|-
-|&nbsp; &nbsp; &nbsp; &nbsp;<math>\lim_{n\rightarrow \infty}\cos(n\pi)\bigg(\frac{1+n}{n}\bigg)^n=\lim_{n\rightarrow \infty} (-1)^n\bigg(\frac{1+n}{n}\bigg)^n.</math>
-|}
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 2: &nbsp;
-|-
-|Now, let
-|-
-|&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
-\displaystyle{y} & = & \displaystyle{\lim_{n\rightarrow \infty} \bigg(\frac{1+n}{n}\bigg)^n.}
-\end{array}</math>
-|-
-|We then take the natural log of both sides to get
-|-
-|&nbsp; &nbsp; &nbsp; &nbsp; <math>\ln y = \ln\bigg(\lim_{n\rightarrow \infty} \bigg(\frac{1+n}{n}\bigg)^n\bigg).</math>
-|-
-|We can interchange limits and continuous functions.
-|-
-|Therefore, we have
-|-
-|
-&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
-\displaystyle{\ln y} & = & \displaystyle{\lim_{n\rightarrow \infty} \ln \bigg(\frac{1+n}{n}\bigg)^n}\\
-&&\\
-& = & \displaystyle{\lim_{n\rightarrow \infty} n\ln\bigg(\frac{1+n}{n}\bigg)}\\
-&&\\
-& = & \displaystyle{\lim_{n\rightarrow \infty} \frac{\ln \bigg(\frac{1+n}{n}\bigg)}{\frac{1}{n}}.}
-\end{array}</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.
-|}
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 3: &nbsp;
-|-
-|Now, we have
-|-
-|
-&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
-\displaystyle{\ln y } & = & \displaystyle{\lim_{n\rightarrow \infty} \frac{\ln \bigg(\frac{1+n}{n}\bigg)}{\frac{1}{n}}}\\
-&&\\
-& = & \displaystyle{\lim_{x\rightarrow \infty} \frac{\ln \bigg(\frac{1+x}{x}\bigg)}{\frac{1}{x}}}\\
-&&\\
-& \overset{L'H}{=} & \displaystyle{\lim_{x\rightarrow \infty} \frac{\frac{x}{1+x}\big(\frac{-1}{x^2}\big)}{\big(-\frac{1}{x^2}\big)}}\\
-&&\\
-& = & \displaystyle{\lim_{x\rightarrow \infty} \frac{x}{1+x}}\\
-&&\\
-& = & \displaystyle{1.}
-\end{array}</math>
-|}
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 4: &nbsp;
-|-
-|Since &nbsp;<math style="vertical-align: -4px">\ln y= 1,</math>&nbsp; we know
-|-
-|&nbsp; &nbsp; &nbsp; &nbsp; <math>y=e.</math>
-|-
-|Since
-|-
-|&nbsp; &nbsp; &nbsp; &nbsp;<math>\lim_{n\rightarrow \infty} \bigg(\frac{1+n}{n}\bigg)^n\neq 0,</math>
-|-
-|we have
-|-
-|
-&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
-\displaystyle{\lim_{n\rightarrow \infty} a_n} & = & \displaystyle{\lim_{n\rightarrow \infty} (-1)^n\bigg(\frac{1+n}{n}\bigg)^n}\\
-&&\\
-& = & \displaystyle{\text{DNE}.}
-\end{array}</math>
-|}
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Final Answer: &nbsp;
-|-
-|&nbsp; &nbsp; '''(a)''' &nbsp; &nbsp;<math>e^{\frac{1}{2}}</math>
-|-
-|&nbsp; &nbsp; '''(b)''' &nbsp; &nbsp;<math>\text{DNE}</math>
-|}
 [[009C_Sample_Final_3|'''<u>Return to Sample Exam</u>''']]

Difference between revisions of "009C Sample Final 3, Problem 1"

Latest revision as of 18:39, 2 December 2017

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