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| | <span class="exam">(b) <math style="vertical-align: -15px">a_n=\bigg(\frac{n}{n+1}\bigg)^n</math> | | <span class="exam">(b) <math style="vertical-align: -15px">a_n=\bigg(\frac{n}{n+1}\bigg)^n</math> |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| + | <hr> |
| − | !Foundations:
| + | [[009C Sample Final 2, Problem 1 Solution|'''<u>Solution</u>''']] |
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| − | |'''L'Hopital's Rule''' | |
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| − | Suppose that <math>\lim_{x\rightarrow \infty} f(x)</math> and <math>\lim_{x\rightarrow \infty} g(x)</math> are both zero or both <math style="vertical-align: -1px">\pm \infty .</math>
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| − | If <math>\lim_{x\rightarrow \infty} \frac{f'(x)}{g'(x)}</math> is finite or <math style="vertical-align: -4px">\pm \infty ,</math>
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| − | then <math>\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:''' | + | [[009C Sample Final 2, Problem 1 Detailed Solution|'''<u>Detailed Solution</u>''']] |
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| − | '''(a)'''
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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, we notice that <math>\lim_{n\rightarrow \infty} \frac{\ln(n)}{\ln(n+1)}</math> has the form <math>\frac{\infty}{\infty}.</math>
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| − | |-
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| − | |So, we can use L'Hopital's Rule. To begin, we write
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| − | | <math>\lim_{n\rightarrow \infty}\frac{\ln(n)}{\ln(n+1)}=\lim_{x\rightarrow \infty} \frac{\ln(x)}{\ln(x+1)}.</math>
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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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| − | |-
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| − | |Now, using L'Hopital's rule, we get
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| − | |-
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| − | | <math>\begin{array}{rcl}
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| − | \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}}}\\
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| − | &&\\
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| − | & = & \displaystyle{\lim_{x\rightarrow \infty} \frac{x+1}{x}}\\
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| − | &&\\
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| − | & = & \displaystyle{\lim_{x\rightarrow \infty} 1+\frac{1}{x}}\\
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| − | &&\\
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| − | & = & \displaystyle{1.}
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| − | \end{array}</math>
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| − | |}
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| − |
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| − | '''(b)'''
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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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| − | |Let
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| − | <math>\begin{array}{rcl}
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| − | \displaystyle{y} & = & \displaystyle{\lim_{n\rightarrow \infty} \bigg(\frac{n}{n+1}\bigg)^n.}
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| − | \end{array}</math>
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| − | |-
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| − | |We then take the natural log of both sides to get
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| − | | <math>\ln y = \ln\bigg(\lim_{n\rightarrow \infty} \bigg(\frac{n}{n+1}\bigg)^n\bigg).</math>
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| − | |}
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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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| − | |-
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| − | |We can interchange limits and continuous functions.
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| − | |Therefore, we have
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| − | |-
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| − | <math>\begin{array}{rcl}
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| − | \displaystyle{\ln y} & = & \displaystyle{\lim_{n\rightarrow \infty} \ln \bigg(\frac{n}{n+1}\bigg)^n}\\
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| − | &&\\
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| − | & = & \displaystyle{\lim_{n\rightarrow \infty} n\ln\bigg(\frac{n}{n+1}\bigg)}\\
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| − | &&\\
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| − | & = & \displaystyle{\lim_{n\rightarrow \infty} \frac{\ln \bigg(\frac{n}{n+1}\bigg)}{\frac{1}{n}}.}
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| − | \end{array}</math>
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| − | |-
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| − | |Now, this limit has the form <math style="vertical-align: -13px">\frac{0}{0}.</math>
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| − | |-
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| − | |Hence, we can use L'Hopital's Rule to calculate this limit.
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| − | |}
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| − |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 3:
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| − | |-
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| − | |Now, we have
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| − | |-
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| − | <math>\begin{array}{rcl}
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| − | \displaystyle{\ln y } & = & \displaystyle{\lim_{n\rightarrow \infty} \frac{\ln \bigg(\frac{n}{n+1}\bigg)}{\frac{1}{n}}}\\
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| − | &&\\
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| − | & = & \displaystyle{\lim_{x\rightarrow \infty} \frac{\ln \bigg(\frac{x}{x+1}\bigg)}{\frac{1}{x}}}\\
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| − | &&\\
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| − | & \overset{L'H}{=} & \displaystyle{\lim_{x\rightarrow \infty} \frac{\frac{x+1}{x}\frac{1}{(x+1)^2}}{\big(-\frac{1}{x^2}\big)}}\\
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| − | &&\\
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| − | & = & \displaystyle{\lim_{x\rightarrow \infty} \frac{-x^2}{x(x+1)}}\\
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| − | &&\\
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| − | & = & \displaystyle{-1.}
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| − | \end{array}</math>
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| − | |}
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| − |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 4:
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| − | |-
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| − | |Since <math>\ln y= -1,</math> we know
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| − | | <math>y=e^{-1}.</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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| − | | '''(a)''' <math>1</math>
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| − | |-
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| − | | '''(b)''' <math>e^{-1}</math>
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| − | |}
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| | [[009C_Sample_Final_2|'''<u>Return to Sample Exam</u>''']] | | [[009C_Sample_Final_2|'''<u>Return to Sample Exam</u>''']] |
