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| | <span class="exam">(c) <math style="vertical-align: -15px">\lim_{x\rightarrow -\infty} \frac{\sqrt{x^2+2}}{2x-1}</math> | | <span class="exam">(c) <math style="vertical-align: -15px">\lim_{x\rightarrow -\infty} \frac{\sqrt{x^2+2}}{2x-1}</math> |
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| | + | [[009A Sample Final 2, Problem 1 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'''
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| − | |-
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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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| − | |-
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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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| − | |-
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| − | |
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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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| − | |}
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| | + | [[009A Sample Final 2, Problem 1 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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| − | |We begin by noticing that we plug in <math style="vertical-align: 0px">x=4</math> into
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| − | |-
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| − | | <math>\frac{\sqrt{x+5}-3}{x-4},</math>
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| − | |-
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| − | |we get <math style="vertical-align: -12px">\frac{0}{0}.</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 multiply the numerator and denominator by the conjugate of the numerator.
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| − | |-
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| − | |Hence, we have
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| − | |-
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| − | | <math>\begin{array}{rcl}
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| − | \displaystyle{\lim_{x\rightarrow 4} \frac{\sqrt{x+5}-3}{x-4}} & = & \displaystyle{\lim_{x\rightarrow 4} \frac{\sqrt{x+5}-3}{x-4}\frac{(\sqrt{x+5}+3)}{(\sqrt{x+5}+3)}}\\
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| − | &&\\
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| − | & = & \displaystyle{\lim_{x\rightarrow 4} \frac{(x+5)-9}{(x-4)(\sqrt{x+5}+3)}}\\
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| − | &&\\
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| − | & = & \displaystyle{\lim_{x\rightarrow 4} \frac{x-4}{(x-4)(\sqrt{x+5}+3)}}\\
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| − | &&\\
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| − | & = & \displaystyle{\lim_{x\rightarrow 4} \frac{1}{\sqrt{x+5}+3}}\\
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| − | &&\\
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| − | & = & \displaystyle{ \frac{1}{\sqrt{9}+3}}\\
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| − | &&\\
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| − | & = & \displaystyle{\frac{1}{6}.}
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| − | \end{array}</math>
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| − | |-
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| − | |
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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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| − | |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 0} \frac{\sin^2 (x)}{3x}} & \overset{L'H}{=} & \displaystyle{\lim_{x\rightarrow 0}\frac{2\sin(x)\cos(x)}{3}.}
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| − | \end{array}</math>
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| − | |-
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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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| − | !Step 2:
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| − | |-
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| − | |Now, we plug in <math style="vertical-align: 0px">x=0</math> to get
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| − | |-
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| − | | <math>\begin{array}{rcl}
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| − | \displaystyle{\lim_{x\rightarrow 0} \frac{\sin^2 (x)}{3x}} & = & \displaystyle{\frac{2\sin(0)\cos(0)}{3}}\\
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| − | &&\\
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| − | & = & \displaystyle{\frac{2(0)(1)}{3}}\\
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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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| − |
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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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| − | |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{\sqrt{x^2+2}}{2x-1}} & = & \displaystyle{\lim_{x\rightarrow -\infty} \frac{\sqrt{x^2(1+\frac{2}{x^2})}}{2x-1}}\\
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| − | &&\\
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| − | & = & \displaystyle{\lim_{x\rightarrow -\infty} \frac{|x|\sqrt{1+\frac{2}{x^2}}}{2x-1}}\\
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| − | &&\\
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| − | & = & \displaystyle{\lim_{x\rightarrow -\infty} \frac{-x\sqrt{1+\frac{2}{x^2}}}{x(2-\frac{1}{x})}}\\
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| − | &&\\
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| − | & = & \displaystyle{\lim_{x\rightarrow -\infty} \frac{-1\sqrt{1+\frac{2}{x^2}}}{(2-\frac{1}{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,
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| − | |-
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| − | | <math>\begin{array}{rcl}
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| − | \displaystyle{\lim_{x\rightarrow -\infty} \frac{\sqrt{x^2+2}}{2x-1}} & = & \displaystyle{\lim_{x\rightarrow -\infty} \frac{-1\sqrt{1+\frac{2}{x^2}}}{(2-\frac{1}{x})} }\\
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| − | &&\\
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| − | & = & \displaystyle{\frac{-\sqrt{1+0}}{(2-0)}}\\
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| − | &&\\
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| − | & = & \displaystyle{\frac{-1}{2}.}
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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>\frac{1}{6}</math>
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| − | |-
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| − | | '''(b)''' <math>0</math>
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| − | |-
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| − | | '''(c)''' <math>\frac{-1}{2}</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>''']] |
