Difference between revisions of "009C Sample Final 2, Problem 1"

Latest revision as of 18:20, 2 December 2017

Test if the following sequences converge or diverge. Also find the limit of each convergent sequence.

(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=\frac{\ln(n)}{\ln(n+1)}}

(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=\bigg(\frac{n}{n+1}\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=\bigg(\frac{n}{n+1}\bigg)^n</math>
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
+<hr>
-!Foundations: &nbsp;
+[[009C Sample Final 2, Problem 1 Solution|'''<u>Solution</u>''']]
-|-
-|'''L'Hopital's Rule'''
-|-
-|
-&nbsp; &nbsp; &nbsp; &nbsp; Suppose that &nbsp;<math>\lim_{x\rightarrow \infty} f(x)</math> &nbsp; and &nbsp;<math>\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>\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>\lim_{x\rightarrow \infty} \frac{f(x)}{g(x)}=\lim_{x\rightarrow \infty} \frac{f'(x)}{g'(x)}.</math>
-|}
-'''Solution:'''
+[[009C Sample Final 2, Problem 1 Detailed Solution|'''<u>Detailed Solution</u>''']]
-'''(a)'''
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 1: &nbsp;
-|-
-|First, we notice that &nbsp;<math>\lim_{n\rightarrow \infty} \frac{\ln(n)}{\ln(n+1)}</math>&nbsp; has the form &nbsp;<math>\frac{\infty}{\infty}.</math>
-|-
-|So, we can use L'Hopital's Rule. To begin, we write
-|-
-|&nbsp; &nbsp; &nbsp; &nbsp;<math>\lim_{n\rightarrow \infty}\frac{\ln(n)}{\ln(n+1)}=\lim_{x\rightarrow \infty} \frac{\ln(x)}{\ln(x+1)}.</math>
-|}
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 2: &nbsp;
-|-
-|Now, using L'Hopital's rule, we get
-|-
-|&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
-\displaystyle{\lim_{n\rightarrow \infty}\frac{\ln(n)}{\ln(n+1)}} & \overset{L'H}{=} & \displaystyle{\lim_{x\rightarrow \infty} \frac{\frac{1}{x}}{\frac{1}{x+1}}}\\
-&&\\
-& = & \displaystyle{\lim_{x\rightarrow \infty} \frac{x+1}{x}}\\
-&&\\
-& = & \displaystyle{\lim_{x\rightarrow \infty} 1+\frac{1}{x}}\\
-&&\\
-& = & \displaystyle{1.}
-\end{array}</math>
-|}
-'''(b)'''
-{| 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(\frac{n}{n+1}\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{n}{n+1}\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(\frac{n}{n+1}\bigg)^n}\\
-&&\\
-& = & \displaystyle{\lim_{n\rightarrow \infty} n\ln\bigg(\frac{n}{n+1}\bigg)}\\
-&&\\
-& = & \displaystyle{\lim_{n\rightarrow \infty} \frac{\ln \bigg(\frac{n}{n+1}\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{n}{n+1}\bigg)}{\frac{1}{n}}}\\
-&&\\
-& = & \displaystyle{\lim_{x\rightarrow \infty} \frac{\ln \bigg(\frac{x}{x+1}\bigg)}{\frac{1}{x}}}\\
-&&\\
-& \overset{L'H}{=} & \displaystyle{\lim_{x\rightarrow \infty} \frac{\frac{x+1}{x}\frac{1}{(x+1)^2}}{\big(-\frac{1}{x^2}\big)}}\\
-&&\\
-& = & \displaystyle{\lim_{x\rightarrow \infty} \frac{-x^2}{x(x+1)}}\\
-&&\\
-& = & \displaystyle{-1.}
-\end{array}</math>
-|}
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 4: &nbsp;
-|-
-|Since &nbsp;<math>\ln y= -1,</math>&nbsp; we know
-|-
-|&nbsp; &nbsp; &nbsp; &nbsp; <math>y=e^{-1}.</math>
-|}
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Final Answer: &nbsp;
-|-
-|&nbsp;&nbsp; '''(a)''' &nbsp; &nbsp; <math>1</math>
-|-
-|&nbsp;&nbsp; '''(b)''' &nbsp; &nbsp; <math>e^{-1}</math>
-|}
 [[009C_Sample_Final_2|'''<u>Return to Sample Exam</u>''']]

Difference between revisions of "009C Sample Final 2, Problem 1"

Latest revision as of 18:20, 2 December 2017

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