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| | <span class="exam">(b) <math style="vertical-align: -12px">\lim_{n\rightarrow \infty} \frac{\ln n}{\ln 3n}</math> | | <span class="exam">(b) <math style="vertical-align: -12px">\lim_{n\rightarrow \infty} \frac{\ln n}{\ln 3n}</math> |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| + | <hr> |
| − | !Foundations:
| + | [[009C Sample Final 1, 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 1, 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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| − | |First, we switch to the limit to <math style="vertical-align: 0px">x</math> so that we can use L'Hopital's rule.
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| − | |So, we have
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| − | <math>\begin{array}{rcl}
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| − | \displaystyle{\lim_{x \rightarrow \infty}\frac{3-2x^2}{5x^2 + x +1}} & \overset{L'H}{=} & \displaystyle{\lim_{x \rightarrow \infty}\frac{-4x}{10x+1}}\\
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| − | &&\\
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| − | & \overset{L'H}{=} & \displaystyle{\frac{-4}{10}}\\
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| − | &&\\
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| − | & = & \displaystyle{-\frac{2}{5}}.
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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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| − | !Step 2:
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| − | |Hence, we have
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| − | <math>\lim_{n\rightarrow \infty} \frac{3-2n^2}{5n^2+n+1}=-\frac{2}{5}.</math>
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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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| − | |Again, we switch to the limit to <math style="vertical-align: 0px">x</math> so that we can use L'Hopital's rule.
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| − | |So, we have
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| − | <math>\begin{array}{rcl}
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| − | \displaystyle{\lim_{x \rightarrow \infty}\frac{\ln x}{\ln(3x)}} & \overset{l'H}{=} & \displaystyle{\lim_{x \rightarrow \infty}\frac{\big(\frac{1}{x}\big)}{\big(\frac{3}{3x}\big)}}\\
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| − | &&\\
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| − | & = & \displaystyle{\lim_{x \rightarrow \infty} 1}\\
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| − | &&\\
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| − | & = & 1.
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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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| − | !Step 2:
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| − | |Hence, we have
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| − | <math>\lim_{n\rightarrow \infty} \frac{\ln n}{\ln 3n}=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 style="vertical-align: -14px">-\frac{2}{5}</math>
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| − | | '''(b)''' <math style="vertical-align: -3px">1</math>
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| − | |}
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| | [[009C_Sample_Final_1|'''<u>Return to Sample Exam</u>''']] | | [[009C_Sample_Final_1|'''<u>Return to Sample Exam</u>''']] |
