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| | ::<math>a_n=\frac{\ln n}{n}</math> | | ::<math>a_n=\frac{\ln n}{n}</math> |
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| | + | [[009C Sample Midterm 1, Problem 1 Solution|'''<u>Solution</u>''']] |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| + | [[009C Sample Midterm 1, Problem 1 Detailed Solution|'''<u>Detailed Solution</u>''']] |
| − | !Foundations:
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| − | |-
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| − | |'''L'Hôpital's Rule, Part 2''' | |
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| − | Let <math style="vertical-align: -5px">f</math> and <math style="vertical-align: -5px">g</math> be differentiable functions on the open interval <math style="vertical-align: -5px">(a,\infty)</math> for some value <math style="vertical-align: -4px">a,</math>
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| − | |-
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| − | | where <math style="vertical-align: -5px">g'(x)\ne 0</math> on <math style="vertical-align: -5px">(a,\infty)</math> and <math style="vertical-align: -18px">\lim_{x\rightarrow \infty} \frac{f(x)}{g(x)}</math> returns either <math style="vertical-align: -15px">\frac{0}{0}</math> or <math style="vertical-align: -15px">\frac{\infty}{\infty}.</math>
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| − | |-
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| − | | Then, <math style="vertical-align: -18px">\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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| − | '''Solution:'''
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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, notice that
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| − | | <math>\lim_{n\rightarrow \infty} \ln n =\infty</math>
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| − | |and
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| − | | <math>\lim_{n\rightarrow \infty} n=\infty.</math>
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| − | |Therefore, the limit has the form <math style="vertical-align: -11px">\frac{\infty}{\infty},</math>
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| − | |-
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| − | |which means that 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 2:
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| − | |First, switch to the variable <math style="vertical-align: 0px">x</math> so that we have functions and
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| − | |-
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| − | |can take derivatives. Thus, using L'Hopital's Rule, we have
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| − | |-
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| − | | <math>\begin{array}{rcl}
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| − | \displaystyle{\lim_{n\rightarrow \infty} \frac{\ln n}{n}} & = & \displaystyle{\lim_{x\rightarrow \infty} \frac{\ln x}{x}}\\
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| − | &&\\
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| − | & \overset{L'H}{=} & \displaystyle{\lim_{x\rightarrow \infty} \frac{\big(\frac{1}{x}\big)}{1}}\\
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| − | &&\\
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| − | & = & \displaystyle{0.}
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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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| − | | The sequence converges. The limit of the sequence is <math style="vertical-align: 0px">0.</math>
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| − | |}
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| | [[009C_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']] | | [[009C_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']] |
