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| | <span class="exam">(c) <math style="vertical-align: -16px">\lim_{x\rightarrow -2} \frac{x^2-x-6}{x^3+8}</math> | | <span class="exam">(c) <math style="vertical-align: -16px">\lim_{x\rightarrow -2} \frac{x^2-x-6}{x^3+8}</math> |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| + | <hr> |
| − | !Foundations:
| + | [[009A Sample Final 3, Problem 7 Solution|'''<u>Solution</u>''']] |
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| − | |'''L'Hôpital's Rule''' | |
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| − | | Suppose that <math style="vertical-align: -11px">\lim_{x\rightarrow \infty} f(x)</math> and <math style="vertical-align: -11px">\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 style="vertical-align: -19px">\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 style="vertical-align: -19px">\lim_{x\rightarrow \infty} \frac{f(x)}{g(x)}\,=\,\lim_{x\rightarrow \infty} \frac{f'(x)}{g'(x)}.</math>
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| − | '''Solution:''' | + | [[009A Sample Final 3, Problem 7 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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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 2:
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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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| − | |We proceed using L'Hôpital's Rule. So, we have
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| − | <math>\begin{array}{rcl}
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| − | \displaystyle{\lim_{x\rightarrow \pi} \frac{\sin (x)}{\pi-x}} & \overset{L'H}{=} & \displaystyle{\lim_{x\rightarrow \pi}\frac{\cos(x)}{-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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| − | |Now, we plug in <math style="vertical-align: 0px">x=\pi</math> to get
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| − | | <math>\begin{array}{rcl}
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| − | \displaystyle{\lim_{x\rightarrow \pi} \frac{\sin (x)}{\pi-x}} & = & \displaystyle{\frac{\cos(\pi)}{-1}}\\
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| − | &&\\
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| − | & = & \displaystyle{\frac{-1}{-1}}\\
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| − | &&\\
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| − | & = & \displaystyle{1.}
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| − | \end{array}</math>
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| − | '''(c)'''
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 1:
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| − | |We begin by factoring the numerator and denominator. We have
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| − | <math>\lim_{x\rightarrow -2} \frac{x^2-x-6}{x^3+8}\,=\,\lim_{x\rightarrow -2}\frac{(x+2)(x-3)}{(x+2)(x^2-2x+4)}.</math>
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| − | |So, we can cancel <math style="vertical-align: -2px">x+2</math> in the numerator and denominator. Thus, we have
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| − | <math>\lim_{x\rightarrow -2} \frac{x^2-x-6}{x^3+8}\,=\,\lim_{x\rightarrow -2}\frac{x-3}{x^2-2x+4}.</math>
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 2:
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| − | |Now, we can just plug in <math style="vertical-align: -1px">x=-2</math> to get
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| − | | <math>\begin{array}{rcl}
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| − | \displaystyle{\lim_{x\rightarrow -2} \frac{x^2-x-6}{x^3+8}} & = & \displaystyle{\frac{-2-3}{(-2)^2-2(-2)+4}}\\
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| − | &&\\
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| − | & = & \displaystyle{\frac{-5}{12}.}
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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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| − | |'''(a)'''
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| − | | '''(b)''' <math>1</math>
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| − | | '''(c)''' <math>\frac{-5}{12}</math>
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| − | |}
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| | [[009A_Sample_Final_3|'''<u>Return to Sample Exam</u>''']] | | [[009A_Sample_Final_3|'''<u>Return to Sample Exam</u>''']] |
