|
|
| (One intermediate revision by the same user not shown) |
| Line 7: |
Line 7: |
| | <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> |
| | | | |
| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| + | <hr> |
| − | !Foundations:
| + | [[009A Sample Final 3, Problem 7 Solution|'''<u>Solution</u>''']] |
| − | |-
| |
| − | |'''L'Hôpital's Rule, Part 1''' | |
| − | |-
| |
| − | |
| |
| − | 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
| |
| − | |-
| |
| − | | 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>
| |
| − | |-
| |
| − | | 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>
| |
| − | |}
| |
| | | | |
| | | | |
| − | '''Solution:''' | + | [[009A Sample Final 3, Problem 7 Detailed Solution|'''<u>Detailed Solution</u>''']] |
| | | | |
| − | '''(a)'''
| |
| | | | |
| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| |
| − | !Step 1:
| |
| − | |-
| |
| − | |We begin by noticing that we plug in <math style="vertical-align: 0px">x=0</math> into
| |
| − | |-
| |
| − | | <math>\frac{x}{3-\sqrt{9-x}},</math>
| |
| − | |-
| |
| − | |we get <math style="vertical-align: -12px">\frac{0}{0}.</math>
| |
| − | |}
| |
| − |
| |
| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| |
| − | !Step 2:
| |
| − | |-
| |
| − | |Now, we multiply the numerator and denominator by the conjugate of the denominator.
| |
| − | |-
| |
| − | |Hence, we have
| |
| − | |-
| |
| − | | <math>\begin{array}{rcl}
| |
| − | \displaystyle{\lim_{x\rightarrow 0} \frac{x}{3-\sqrt{9-x}}} & = & \displaystyle{\lim_{x\rightarrow 0} \frac{x}{3-\sqrt{9-x}}\frac{(3+\sqrt{9+x})}{(3+\sqrt{9+x})}}\\
| |
| − | &&\\
| |
| − | & = & \displaystyle{\lim_{x\rightarrow 0} \frac{x(3+\sqrt{9+x})}{9-(9+x)}}\\
| |
| − | &&\\
| |
| − | & = & \displaystyle{\lim_{x\rightarrow 0} \frac{x(3+\sqrt{9+x})}{-x}}\\
| |
| − | &&\\
| |
| − | & = & \displaystyle{\lim_{x\rightarrow 0} \frac{3+\sqrt{9+x}}{-1}}\\
| |
| − | &&\\
| |
| − | & = & \displaystyle{ \frac{3+\sqrt{9}}{-1}}\\
| |
| − | &&\\
| |
| − | & = & \displaystyle{-\frac{6}{1}}\\
| |
| − | &&\\
| |
| − | & = & \displaystyle{-6.}
| |
| − | \end{array}</math>
| |
| − | |-
| |
| − | |
| |
| − | |}
| |
| − |
| |
| − | '''(b)'''
| |
| − |
| |
| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| |
| − | !Step 1:
| |
| − | |-
| |
| − | |We proceed using L'Hôpital's Rule. So, we have
| |
| − | |-
| |
| − | |
| |
| − | <math>\begin{array}{rcl}
| |
| − | \displaystyle{\lim_{x\rightarrow \pi} \frac{\sin (x)}{\pi-x}} & \overset{L'H}{=} & \displaystyle{\lim_{x\rightarrow \pi}\frac{\cos(x)}{-1}.}
| |
| − | \end{array}</math>
| |
| − | |}
| |
| − |
| |
| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| |
| − | !Step 2:
| |
| − | |-
| |
| − | |Now, we plug in <math style="vertical-align: 0px">x=\pi</math> to get
| |
| − | |-
| |
| − | | <math>\begin{array}{rcl}
| |
| − | \displaystyle{\lim_{x\rightarrow \pi} \frac{\sin (x)}{\pi-x}} & = & \displaystyle{\frac{\cos(\pi)}{-1}}\\
| |
| − | &&\\
| |
| − | & = & \displaystyle{\frac{-1}{-1}}\\
| |
| − | &&\\
| |
| − | & = & \displaystyle{1.}
| |
| − | \end{array}</math>
| |
| − | |}
| |
| − |
| |
| − | '''(c)'''
| |
| − |
| |
| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| |
| − | !Step 1:
| |
| − | |-
| |
| − | |We begin by factoring the numerator and denominator. We have
| |
| − | |-
| |
| − | |
| |
| − | <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>
| |
| − | |-
| |
| − | |So, we can cancel <math style="vertical-align: -2px">x+2</math> in the numerator and denominator. Thus, we have
| |
| − | |-
| |
| − | |
| |
| − | <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>
| |
| − | |}
| |
| − |
| |
| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| |
| − | !Step 2:
| |
| − | |-
| |
| − | |Now, we can just plug in <math style="vertical-align: -1px">x=-2</math> to get
| |
| − | |-
| |
| − | | <math>\begin{array}{rcl}
| |
| − | \displaystyle{\lim_{x\rightarrow -2} \frac{x^2-x-6}{x^3+8}} & = & \displaystyle{\frac{-2-3}{(-2)^2-2(-2)+4}}\\
| |
| − | &&\\
| |
| − | & = & \displaystyle{-\frac{5}{12}.}
| |
| − | \end{array}</math>
| |
| − | |}
| |
| − |
| |
| − |
| |
| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| |
| − | !Final Answer:
| |
| − | |-
| |
| − | | '''(a)''' <math>-6</math>
| |
| − | |-
| |
| − | | '''(b)''' <math>1</math>
| |
| − | |-
| |
| − | | '''(c)''' <math>-\frac{5}{12}</math>
| |
| − | |}
| |
| | [[009A_Sample_Final_3|'''<u>Return to Sample Exam</u>''']] | | [[009A_Sample_Final_3|'''<u>Return to Sample Exam</u>''']] |
