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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>''']] | |
In each part, compute the limit. If the limit is infinite, be sure to specify positive or negative infinity.
(a) 
(b) 
(c) Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{x\rightarrow -\infty} \frac{3x}{\sqrt{4x^2+x+5}}}
Solution
Detailed Solution
Return to Sample Exam