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| | <span class="exam">(c) <math style="vertical-align: -14px">\lim_{x\rightarrow -\infty} \frac{\sqrt{9x^6-x}}{3x^3+4x}</math> | | <span class="exam">(c) <math style="vertical-align: -14px">\lim_{x\rightarrow -\infty} \frac{\sqrt{9x^6-x}}{3x^3+4x}</math> |
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| | + | <hr> |
| | + | [[009A Sample Final 3, 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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| − | | '''1.''' If <math style="vertical-align: -12px">\lim_{x\rightarrow a} g(x)\neq 0,</math> we have
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| − | |-
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| − | | <math>\lim_{x\rightarrow a} \frac{f(x)}{g(x)}=\frac{\displaystyle{\lim_{x\rightarrow a} f(x)}}{\displaystyle{\lim_{x\rightarrow a} g(x)}}.</math>
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| − | |-
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| − | | '''2.''' <math style="vertical-align: -14px">\lim_{x\rightarrow 0} \frac{\sin x}{x}=1</math>
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| − | |}
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| | + | [[009A Sample Final 3, 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=0</math> into
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| − | |-
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| − | | <math>\frac{\sin(5x)}{1-\sqrt{1-x}},</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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| − | {| 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 denominator.
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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 0} \frac{\sin(5x)}{1-\sqrt{1-x}}} & = & \displaystyle{\lim_{x\rightarrow 0} \frac{\sin(5x)}{1-\sqrt{1-x}} \bigg(\frac{1+\sqrt{1-x}}{1+\sqrt{1-x}}\bigg)}\\
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| − | &&\\
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| − | & = & \displaystyle{\lim_{x\rightarrow 0} \frac{\sin(5x)(1+\sqrt{1-x})}{x}}\\
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| − | &&\\
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| − | & = & \displaystyle{\lim_{x\rightarrow 0} \frac{\sin(5x)}{x}(1+\sqrt{1-x})}\\
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| − | &&\\
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| − | & = & \displaystyle{\bigg(\lim_{x\rightarrow 0} \frac{\sin(5x)}{x}\bigg) \lim_{x\rightarrow 0}(1+\sqrt{1-x})}\\
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| − | &&\\
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| − | & = & \displaystyle{\bigg(5\lim_{x\rightarrow 0} \frac{\sin(5x)}{5x}\bigg) (2)}\\
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| − | &&\\
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| − | & = & \displaystyle{5(1)(2)}\\
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| − | &&\\
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| − | & = & \displaystyle{10.}
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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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| − | |Since <math style="vertical-align: -12px">\lim_{x\rightarrow 8} 3 =3\ne 0,</math>
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| − | |-
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| − | |we have
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| − | | <math>\begin{array}{rcl}
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| − | \displaystyle{-2} & = & \displaystyle{\lim _{x\rightarrow 8} \bigg[\frac{xf(x)}{3}\bigg]}\\
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| − | &&\\
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| − | & = & \displaystyle{\frac{\displaystyle{\lim_{x\rightarrow 8} xf(x)}}{\displaystyle{\lim_{x\rightarrow 8} 3}}}\\
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| − | &&\\
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| − | & = & \displaystyle{\frac{\displaystyle{\lim_{x\rightarrow 8} xf(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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| − | |If we multiply both sides of the last equation by <math>3,</math> we get
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| − | |-
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| − | | <math>-6=\lim_{x\rightarrow 8} xf(x).</math>
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| − | |-
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| − | |Now, using properties of limits, we have
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| − | |-
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| − | | <math>\begin{array}{rcl}
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| − | \displaystyle{-6} & = & \displaystyle{\bigg(\lim_{x\rightarrow 8} x\bigg)\bigg(\lim_{x\rightarrow 8}f(x)\bigg)}\\
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| − | &&\\
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| − | & = & \displaystyle{8\lim_{x\rightarrow 8} f(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 3:
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| − | |-
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| − | |Solving for <math style="vertical-align: -12px">\lim_{x\rightarrow 8} f(x)</math> in the last equation,
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| − | |-
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| − | |we get
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| − | |-
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| − | <math> \lim_{x\rightarrow 8} f(x)=-\frac{3}{4}.</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 write
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| − | |-
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| − | | <math>\begin{array}{rcl}
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| − | \displaystyle{\lim_{x\rightarrow -\infty} \frac{\sqrt{9x^6-x}}{3x^3+4x}} & = & \displaystyle{\lim_{x\rightarrow -\infty} \frac{\sqrt{9x^6-x}}{3x^3+4x}\frac{\big(\frac{1}{x^3}\big)}{\big(\frac{1}{x^3}\big)}}\\
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| − | &&\\
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| − | & = & \displaystyle{\lim_{x\rightarrow -\infty} \frac{\sqrt{9-\frac{1}{x^5}}}{3+\frac{4}{x^2}}.}
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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{\sqrt{9x^6-x}}{3x^3+4x}} & = & \displaystyle{\frac{\lim_{x\rightarrow -\infty} \sqrt{9-\frac{1}{x^5}}}{\lim_{x\rightarrow -\infty}3+\frac{4}{x^2}}}\\
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| − | &&\\
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| − | & = & \displaystyle{\frac{\sqrt{9}}{3}}\\
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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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| − | !Final Answer:
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| − | |-
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| − | | '''(a)''' <math>10</math>
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| − | |-
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| − | | '''(b)''' <math>-\frac{3}{4}</math>
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| − | |-
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| − | | '''(c)''' <math>1</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>''']] |
