009C Sample Midterm 3, Problem 1 - Revision history

MathAdmin at 16:41, 28 November 2017

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MathAdmin at 00:39, 4 November 2015

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MathAdmin at 22:24, 26 April 2015

2015-04-26T22:24:15Z

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2015-04-26T18:50:01Z

Created page with "<span class="exam">Test if the following sequence <math style="vertical-align: -10%">{a_n}</math> converges or diverges. If it converges, also find the limit of the sequence...."

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<span class="exam">Test if the following sequence <math style="vertical-align: -10%">{a_n}</math> converges or diverges. If it converges, also find the limit of the sequence.

::<math>a_{n}=\left(\frac{n-7}{n}\right)^{1/n}.</math>

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
! Foundations:  
|-
|This a common question, and is related to the fact that
|-
|
::<math style="vertical-align: 0%">\lim_{x\rightarrow\infty}\left(1+\frac{\alpha}{x}\right)^{x}\ =\ e^{\alpha}.</math>
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|In such a limit, the argument <math style="vertical-align: -22%">1+\alpha /x</math> tends to one as <math style="vertical-align: 0%">x</math> gets large, while we are raising that argument to an increasing power. Neither one really "wins", so we end up with a finite limit that is neither zero nor infinity.
|-
|On the other hand, in the exam problem the argument <math style="vertical-align: -22%">(n-7)/n</math> is always smaller than one, but tends to one as <math style="vertical-align: 0%">n</math> gets large, while the exponent <math style="vertical-align: -25%">1/n</math> tends to zero. These do not disagree, so the limit should be one, but we need to prove it.
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|Any time you have a function raised to a function, we need to use natural log and take advantage of the log rule:
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::<math style="vertical-align: 0%">\ln\left(a^{b}\right)=b\ln(a).</math>
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|For example, to find <math style="vertical-align: -78%">\lim_{x\rightarrow\infty}\left(1-\frac{1}{x}\right)^{x}</math>, you could begin by saying: Let <math style="vertical-align: -75%">L=\lim_{x\rightarrow\infty}\left(1-\frac{1}{x}\right)^{x}.</math>
Then
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::<math style="vertical-align: 0%">\ln L=\ln\left[\lim_{x\rightarrow\infty}\left(1-\frac{1}{x}\right)^{x}\right]=\lim_{x\rightarrow\infty}\ln\left[\left(1-\frac{1}{x}\right)^{x}\right],</math>

where we are allowed to pass the log through the limit because natural log is continuous. But by log rules,
|-
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::<math style="vertical-align: 0%">\ln\left[\left(1-\frac{1}{x}\right)^{x}\right]=x\ln\left(1-\frac{1}{x}\right).</math>
|-
|Thus
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::<math style="vertical-align: 0%">\begin{array}{rcl}
\lim_{x\rightarrow\infty}\ln\left[\left(1-\frac{1}{x}\right)^{x}\right] & = & \lim_{x\rightarrow\infty}x\ln\left(1-\frac{1}{x}\right),\\
& = & \lim_{x\rightarrow\infty}\frac{\ln\left(1-\frac{1}{x}\right)}{\frac{1}{x}}\\
& \overset{l'H}{=} & \lim_{x\rightarrow\infty}\frac{\frac{x}{x-1}\cdot\left(-\frac{1}{x}\right)'}{\left(\frac{1}{x}\right)'}\\
& = & -1.
\end{array}</math>

Note that <math style="vertical-align: -100%">\lim_{x\rightarrow\infty}\frac{\ln\left(1-\frac{1}{x}\right)}{\frac{1}{x}}=\frac{0}{0},</math>
 so we can apply l'Hôpital's rule. Finally, since <math style="vertical-align: -60%">\ln L=-1,\,\,L=\frac{1}{e}.</math>
|-
|Again, such a technique is not required for this particular problem, as the exponent tends to zero. But the technique is common enough on exams to justify providing an example.
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!'''Solution:''' 
|-
|Following the procedure outlined in Foundations, let <math style="vertical-align: -75%">L=\lim_{n\rightarrow\infty}\left(\frac{n-7}{n}\right)^{1/n}.</math> Then
|-
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::<math style="vertical-align: 0%">\begin{array}{rcl}
\ln L & = & \ln\left(\lim_{n\rightarrow\infty}\left[\left(\frac{n-7}{n}\right)^{1/n}\right]\right)\\
& = & \lim_{n\rightarrow\infty}\ln\left[\left(\frac{n-7}{n}\right)^{1/n}\right]\\
& & \lim_{n\rightarrow\infty}\left[\frac{1}{n}\cdot\ln\left(\frac{n-7}{n}\right)\right]\\
& = & 0\cdot\ln(1)\\
& = & 0.
\end{array}</math>
|-
|Thus, <math style="vertical-align: -5%">L=e^{0}=1.</math> Also, most teachers would require you to mention that natural log is continuous as justification for passing the limit through it.
|}
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|The limit of the sequence is <math style="vertical-align: -5%">e^{0}=1.</math>
|}

[[009C_Sample_Midterm_3|'''<u>Return to Sample Exam</u>''']]

@@ Line 1: / Line 1: @@
-<span class="exam">Test if the following sequence <math  style="vertical-align: -10%">{a_n}</math> converges or diverges. If it converges, also find the limit of the sequence.
+<span class="exam">Test if the following sequence <math  style="vertical-align: -10%">{a_n}</math> converges or diverges.
-::<math>a_{n}=\left(\frac{n-7}{n}\right)^{1/n}.</math>
+<span class="exam">If it converges, also find the limit of the sequence.
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
+::<math>a_{n}=\left(\frac{n-7}{n}\right)^{\frac{1}{n}}</math>
-where we are allowed to pass the log through the limit because natural log is continuous. But by log rules,
+<hr>
-Note that <math style="vertical-align: -100%">\lim_{x\rightarrow\infty}\frac{\ln\left(1-\frac{1}{x}\right)}{\frac{1}{x}}=\frac{0}{0},</math>
+[[009C Sample Midterm 3, Problem 1 Solution|'''<u>Solution</u>''']]
-&thinsp;so we can apply l'H&ocirc;pital's rule. Finally, since <math style="vertical-align: -60%">\ln L=-1,\,\,L=\frac{1}{e}.</math>
-|-
-|Again, such a technique is not required for this particular problem, as the exponent tends to zero.  But the technique is common enough on exams to justify providing an example.
+[[009C Sample Midterm 3, Problem 1 Detailed Solution|'''<u>Detailed Solution</u>''']]
 [[009C_Sample_Midterm_3|'''<u>Return to Sample Exam</u>''']]

@@ Line 65: / Line 65: @@
 \end{array}</math>
 |-
-|<br>Thus, <math style="vertical-align: -5%">L=e^{0}=1.</math>   Also, most teachers would require you to mention that natural log is continuous as justification for passing the limit through it.
+|<br>Thus, <math style="vertical-align: -1px">L=e^{0}=1.</math>   Also, most teachers would require you to mention that natural log is continuous as justification for passing the limit through it.
 |}
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"

@@ Line 54: / Line 54: @@
+|
 ::<math style="vertical-align: 0%">\begin{array}{rcl}
-\ln L & = & \ln\left(\lim_{n\rightarrow\infty}\left[\left(\frac{n-7}{n}\right)^{1/n}\right]\right)\\
+\ln L & = & \displaystyle{ \ln\left(\lim_{n\rightarrow\infty}\left[\left(\frac{n-7}{n}\right)^{1/n}\right]\right)}\\
-  & = & \lim_{n\rightarrow\infty}\ln\left[\left(\frac{n-7}{n}\right)^{1/n}\right]\\
+  \\
- &  & \lim_{n\rightarrow\infty}\left[\frac{1}{n}\cdot\ln\left(\frac{n-7}{n}\right)\right]\\
+& = &\displaystyle{ \lim_{n\rightarrow\infty}\ln\left[\left(\frac{n-7}{n}\right)^{1/n}\right]}\\
- & = & 0\cdot\ln(1)\\
+\\
- & = & 0.
+&  & \displaystyle{ \lim_{n\rightarrow\infty}\left[\frac{1}{n}\cdot\ln\left(\frac{n-7}{n}\right)\right]}\\
 \end{array}</math>
 |-
-|Thus, <math style="vertical-align: -5%">L=e^{0}=1.</math>   Also, most teachers would require you to mention that natural log is continuous as justification for passing the limit through it.
+|<br>Thus, <math style="vertical-align: -5%">L=e^{0}=1.</math>   Also, most teachers would require you to mention that natural log is continuous as justification for passing the limit through it.
 |}
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"

009C Sample Midterm 3, Problem 1 - Revision history

MathAdmin at 16:41, 28 November 2017

MathAdmin at 00:39, 4 November 2015

MathAdmin at 22:24, 26 April 2015

MathAdmin: Created page with "Test if the following sequence {a_n} converges or diverges. If it converges, also find the limit of the sequence...."

MathAdmin: Created page with "Test if the following sequence ${a_n}$ converges or diverges. If it converges, also find the limit of the sequence...."