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| | <span class="exam">(c) <math style="vertical-align: -14px">\lim_{x\rightarrow -\infty} \frac{3x}{\sqrt{4x^2+x+5}}</math> | | <span class="exam">(c) <math style="vertical-align: -14px">\lim_{x\rightarrow -\infty} \frac{3x}{\sqrt{4x^2+x+5}}</math> |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| + | <hr> |
| − | !Foundations:
| + | [[009A Sample Midterm 2, Problem 1 Solution|'''<u>Solution</u>''']] |
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| − | |'''L'Hôpital's Rule, Part 1'''
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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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| − | | 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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| − | | 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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| − | '''Solution:''' | + | [[009A Sample Midterm 2, Problem 1 Detailed Solution|'''<u>Detailed Solution</u>''']] |
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| − | '''(a)'''
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| + | [[009A_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']] |
| − | !Step 1:
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| − | |We begin by factoring the numerator. We have
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| − | <math>\lim_{x\rightarrow -3} \frac{x^3-9x}{6+2x}\,=\,\lim_{x\rightarrow -3}\frac{x(x-3)(x+3)}{2(x+3)}.</math>
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| − | |So, we can cancel <math style="vertical-align: -2px">x+3</math> in the numerator and denominator. Thus, we have
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| − | <math>\lim_{x\rightarrow -3} \frac{x^3-9x}{6+2x}\,=\,\lim_{x\rightarrow -3}\frac{x(x-3)}{2}.</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 can just plug in <math style="vertical-align: -1px">x=-3</math> to get
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| − | <math>\begin{array}{rcl}
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| − | \displaystyle{\lim_{x\rightarrow -3} \frac{x^3-9x}{6+2x}} & = & \displaystyle{\frac{(-3)(-3-3)}{2}}\\
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| − | &&\\
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| − | & = & \displaystyle{\frac{18}{2}}\\
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| − | &&\\
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| − | & = & \displaystyle{9.}
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| − | \end{array}</math>
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| − | |}
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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 0^+} \frac{\sin (2x)}{x^2}} & = & \displaystyle{\lim_{x\rightarrow 0^+}\frac{2\cos(2x)}{2x}}\\
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| − | &&\\
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| − | & = & \displaystyle{\lim_{x\rightarrow 0^+}\frac{\cos(2x)}{x}.}\\
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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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| − | |This limit is <math>\infty.</math>
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| − | |}
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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 have
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| − | <math>\lim_{x\rightarrow -\infty} \frac{3x}{\sqrt{4x^2+x+5}}=\lim_{x\rightarrow -\infty} \frac{3x}{\sqrt{x^2(4+\frac{1}{x}+\frac{5}{x^2}})}.</math>
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| − | |Since we are looking at the limit as <math style="vertical-align: 0px">x</math> goes to negative infinity, we have <math style="vertical-align: -2px">\sqrt{x^2}=-x.</math>
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| − | |So, we have
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| − | <math>\lim_{x\rightarrow -\infty} \frac{3x}{\sqrt{4x^2+x+5}}\,=\,\lim_{x\rightarrow -\infty} \frac{3x}{-x\sqrt{4+\frac{1}{x}+\frac{5}{x^2}}}.</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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| − | | We simplify to get
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| − | <math>\begin{array}{rcl}
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| − | \displaystyle{\lim_{x\rightarrow -\infty} \frac{3x}{\sqrt{4x^2+x+5}}} & = & \displaystyle{\lim_{x\rightarrow -\infty} \frac{-3}{\sqrt{4+\frac{1}{x}+\frac{5}{x^2}}}}\\
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| − | &&\\
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| − | & = & \displaystyle{-\frac{3}{\sqrt{4}}}\\
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| − | &&\\
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| − | & = & \displaystyle{-\frac{3}{2}.}
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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)''' <math style="vertical-align: 0px">9</math>
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| − | | '''(b)''' <math style="vertical-align: 0px">\infty</math>
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| − | | '''(c)''' <math style="vertical-align: -15px">-\frac{3}{2}</math>
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| − | |}
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| − | [[009A_Sample_Final_1|'''<u>Return to Sample Exam</u>''']] | |
