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| | <span class="exam">(c) <math style="vertical-align: -15px">\lim_{x\rightarrow 1} \frac{x^3-1}{x^{10}-1}</math> | | <span class="exam">(c) <math style="vertical-align: -15px">\lim_{x\rightarrow 1} \frac{x^3-1}{x^{10}-1}</math> |
| | + | <hr> |
| | + | [[009A Sample Final 2, Problem 8 Solution|'''<u>Solution</u>''']] |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Foundations:
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| − | |-
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| − | |'''L'Hôpital's Rule, Part 1'''
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| − | |-
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| − | Let <math style="vertical-align: -12px">\lim_{x\rightarrow c}f(x)=0</math> and <math style="vertical-align: -12px">\lim_{x\rightarrow c}g(x)=0,</math> where <math style="vertical-align: -5px">f</math> and <math style="vertical-align: -5px">g</math> are differentiable functions
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| − | |-
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| − | | on an open interval <math style="vertical-align: 0px">I</math> containing <math style="vertical-align: -5px">c,</math> and <math style="vertical-align: -5px">g'(x)\ne 0</math> on <math style="vertical-align: 0px">I</math> except possibly at <math style="vertical-align: 0px">c.</math>
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| − | |-
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| − | | Then, <math style="vertical-align: -18px">\lim_{x\rightarrow c} \frac{f(x)}{g(x)}=\lim_{x\rightarrow c} \frac{f'(x)}{g'(x)}.</math>
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| − | |}
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| | + | [[009A Sample Final 2, Problem 8 Detailed Solution|'''<u>Detailed Solution</u>''']] |
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| − | '''Solution:'''
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| − | '''(a)'''
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| − |
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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 have
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| − | |-
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| − | | <math>\begin{array}{rcl}
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| − | \displaystyle{\lim_{x\rightarrow \infty} \frac{x^{-1}+x}{1+\sqrt{1+x}}} & = & \displaystyle{\lim_{x\rightarrow \infty}\frac{\frac{1}{x}+x}{1+\sqrt{1+x}}}\\
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| − | &&\\
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| − | & = & \displaystyle{\lim_{x\rightarrow \infty}\frac{\frac{1}{x}+x}{1+\sqrt{1+x}} \frac{\big(\frac{1}{\sqrt{x}}\big)}{\big(\frac{1}{\sqrt{x}}\big)}}\\
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| − | &&\\
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| − | & = & \displaystyle{\lim_{x\rightarrow \infty} \frac{\frac{1}{x^{3/2}}+\sqrt{x}}{\frac{1}{\sqrt{x}}+\sqrt{\frac{1}{x}+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 2:
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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{\lim_{x\rightarrow \infty} \frac{x^{-1}+x}{1+\sqrt{1+x}}} & = & \displaystyle{\lim_{x\rightarrow \infty} \frac{\frac{1}{x^{3/2}}+\sqrt{x}}{\frac{1}{\sqrt{x}}+\sqrt{\frac{1}{x}+1}}}\\
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| − | &&\\
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| − | & = & \displaystyle{\frac{\lim_{x\rightarrow \infty}\big(\frac{1}{x^{3/2}}+\sqrt{x}\big)}{\lim_{x\rightarrow \infty}\big(\frac{1}{\sqrt{x}}+\sqrt{\frac{1}{x}+1}\big)}}\\
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| − | &&\\
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| − | & = & \displaystyle{\frac{\lim_{x\rightarrow \infty}\frac{1}{x^{3/2}}+\lim_{x\rightarrow \infty}\sqrt{x}}{\lim_{x\rightarrow \infty}\frac{1}{\sqrt{x}}+\lim_{x\rightarrow \infty}\sqrt{\frac{1}{x}+1}}}\\
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| − | &&\\
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| − | & = & \displaystyle{\frac{0+\lim_{x\rightarrow \infty}\sqrt{x}}{0+1}}\\
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| − | &&\\
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| − | & = & \displaystyle{\infty.}
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| − | \end{array}</math>
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| − | |}
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| − |
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| − | '''(b)'''
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| − |
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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 write
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| − | |-
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| − | | <math>\begin{array}{rcl}
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| − | \displaystyle{\lim_{x\rightarrow 0} \frac{\sin x}{\cos x-1}} & = & \displaystyle{\lim_{x\rightarrow 0} \frac{\sin x}{\cos x-1}\frac{(\cos x+1)}{(\cos x+1)}}\\
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| − | &&\\
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| − | & = & \displaystyle{\lim_{x\rightarrow 0} \frac{\sin x (\cos x+1)}{\cos^2x-1}}\\
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| − | &&\\
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| − | & = & \displaystyle{\lim_{x\rightarrow 0} \frac{\sin x(\cos x+1)}{-\sin^2 x}}\\
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| − | &&\\
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| − | & = & \displaystyle{\lim_{x\rightarrow 0} \frac{\cos x+1}{-\sin x}.}
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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 2:
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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{\lim_{x\rightarrow 0^+} \frac{\sin x}{\cos x-1}} & = & \displaystyle{\lim_{x\rightarrow 0^+} \frac{\cos x+1}{-\sin x}}\\
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| − | &&\\
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| − | & = & \displaystyle{-\infty}
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| − | \end{array}</math>
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| − | |-
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| − | |and
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| − | |-
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| − | | <math>\begin{array}{rcl}
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| − | \displaystyle{\lim_{x\rightarrow 0^-} \frac{\sin x}{\cos x-1}} & = & \displaystyle{\lim_{x\rightarrow 0^-} \frac{\cos x+1}{-\sin x}}\\
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| − | &&\\
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| − | & = & \displaystyle{\infty.}
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| − | \end{array}</math>
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| − | |-
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| − | |Therefore,
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| − | |-
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| − | | <math>\lim_{x\rightarrow 0} \frac{\sin x}{\cos x-1}=\text{DNE}.</math>
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| − | |}
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| − |
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| − | '''(c)'''
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| − |
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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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| − | |We proceed using L'Hôpital's Rule. So, we have
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| − | |-
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| − | |
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| − | <math>\begin{array}{rcl}
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| − | \displaystyle{\lim_{x\rightarrow 1} \frac{x^3-1}{x^{10}-1}} & \overset{L'H}{=} & \displaystyle{\lim_{x\rightarrow 1}\frac{3x^2}{10x^9}.}
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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 2:
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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{\lim_{x\rightarrow 1} \frac{x^3-1}{x^{10}-1}} & \overset{L'H}{=} & \displaystyle{\lim_{x\rightarrow 1}\frac{3x^2}{10x^9}}\\
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| − | &&\\
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| − | & = & \displaystyle{\frac{3(1)^2}{10(1)^9}}\\
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| − | &&\\
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| − | & = & \displaystyle{\frac{3}{10}.}
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| − | \end{array}</math>
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| − | |}
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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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| − | !Final Answer:
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| − | |-
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| − | | '''(a)''' <math>\infty</math>
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| − | |-
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| − | | '''(b)''' <math>\text{DNE}</math>
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| − | |-
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| − | | '''(c)''' <math>\frac{3}{10}</math>
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| − | |}
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| | [[009A_Sample_Final_2|'''<u>Return to Sample Exam</u>''']] | | [[009A_Sample_Final_2|'''<u>Return to Sample Exam</u>''']] |
