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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. | + | <span class="exam">Test if the following sequence <math style="vertical-align: -10%">{a_n}</math> converges or diverges. |
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| − | ::<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. |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| + | ::<math>a_{n}=\left(\frac{n-7}{n}\right)^{\frac{1}{n}}</math> |
| − | ! Foundations:
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| − | |This a common question, and is related to the fact that
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| − | ::<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.
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| − | |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>
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| − | 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>
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| − | 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>
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| − | |Thus
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| − | ::<math style="vertical-align: 0%">\begin{array}{rcl}
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| − | \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),\\
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| − | & = & \lim_{x\rightarrow\infty}\frac{\ln\left(1-\frac{1}{x}\right)}{\frac{1}{x}}\\
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| − | & \overset{l'H}{=} & \lim_{x\rightarrow\infty}\frac{\frac{x}{x-1}\cdot\left(-\frac{1}{x}\right)'}{\left(\frac{1}{x}\right)'}\\
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| − | & = & -1.
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| − | \end{array}</math>
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| − | 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>''']] |
| − |  so we can apply l'Hôpital's rule. Finally, since <math style="vertical-align: -60%">\ln L=-1,\,\,L=\frac{1}{e}.</math>
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| − | |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>''']] |
| − | |}
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !'''Solution:'''
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| − | |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}
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| − | \ln L & = & \ln\left(\lim_{n\rightarrow\infty}\left[\left(\frac{n-7}{n}\right)^{1/n}\right]\right)\\
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| − | & = & \lim_{n\rightarrow\infty}\ln\left[\left(\frac{n-7}{n}\right)^{1/n}\right]\\
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| − | & & \lim_{n\rightarrow\infty}\left[\frac{1}{n}\cdot\ln\left(\frac{n-7}{n}\right)\right]\\
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| − | & = & 0\cdot\ln(1)\\
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| − | & = & 0.
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| − | \end{array}</math>
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| − | |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.
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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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| − | |The limit of the sequence is <math style="vertical-align: -5%">e^{0}=1.</math>
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| − | |}
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| | [[009C_Sample_Midterm_3|'''<u>Return to Sample Exam</u>''']] | | [[009C_Sample_Midterm_3|'''<u>Return to Sample Exam</u>''']] |
