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| | <span class="exam">In each part, compute the limit. If the limit is infinite, be sure to specify positive or negative infinity. | | <span class="exam">In each part, compute the limit. If the limit is infinite, be sure to specify positive or negative infinity. |
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| − | <span class="exam">a) <math style="vertical-align: -14px">\lim_{x\rightarrow -3} \frac{x^3-9x}{6+2x}</math> | + | <span class="exam">(a) <math style="vertical-align: -14px">\lim_{x\rightarrow -3} \frac{x^3-9x}{6+2x}</math> |
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| − | <span class="exam">b) <math style="vertical-align: -14px">\lim_{x\rightarrow 0^+} \frac{\sin (2x)}{x^2}</math> | + | <span class="exam">(b) <math style="vertical-align: -14px">\lim_{x\rightarrow 0^+} \frac{\sin (2x)}{x^2}</math> |
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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> |
| − | == Temp1 ==
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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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| − | |Recall:
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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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| − | |
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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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| − | '''Solution:''' | + | <hr> |
| | + | [[009A Sample Final 1, Problem 1 Solution|'''<u>Solution</u>''']] |
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| − | == Temp2 ==
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| − | '''(a)'''
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| + | [[009A Sample Final 1, Problem 1 Detailed Solution|'''<u>Detailed Solution</u>''']] |
| − | !Step 1:
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| − | |-
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| − | |We begin by factoring the numerator. We have
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| − | |-
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| − | |
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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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| − | |-
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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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| − | |- | |
| − | |
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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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| − | |-
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| − | |Now, we can just plug in <math style="vertical-align: 0px">x=-3</math> to get
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| − | |-
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| − | |
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| − | ::<math>\lim_{x\rightarrow -3} \frac{x^3-9x}{6+2x}=\frac{(-3)(-3-3)}{2}=\frac{18}{2}=9.</math>
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| − | |}
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| − | == Temp3 ==
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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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| − | |-
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| − | |We proceed using L'Hopital'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 (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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| − |
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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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| − | |This limit is <math>+\infty.</math>
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| − | |}
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| − | == Temp4 ==
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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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| − | | We have
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| − | |-
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| − | |
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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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| − | |-
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| − | |Since we are looking at the limit as <math>x</math> goes to negative infinity, we have <math style="vertical-align: -3px">\sqrt{x^2}=-x.</math>
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| − | |-
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| − | |So, we have
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| − | |-
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| − | |
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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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| − |
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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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| − | | We simplify to get
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| − | |-
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| − | |
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| − | ::<math>\lim_{x\rightarrow -\infty} \frac{3x}{\sqrt{4x^2+x+5}}=\lim_{x\rightarrow -\infty} \frac{-3}{\sqrt{4+\frac{1}{x}+\frac{5}{x^2}}}.</math>
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| − | |-
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| − | |So, we have
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| − | |-
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| − | |
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| − | ::<math>\lim_{x\rightarrow -\infty} \frac{3x}{\sqrt{4x^2+x+5}}=\frac{-3}{\sqrt{4}}=\frac{-3}{2}.</math>
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| − | |}
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| − | == Temp5 ==
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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>9</math>.
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| − | |-
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| − | |'''(b)''' <math>+\infty</math>
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| − | |-
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| − | |'''(c)''' <math>\frac{-3}{2}</math>
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| − | |}
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| | [[009A_Sample_Final_1|'''<u>Return to Sample Exam</u>''']] | | [[009A_Sample_Final_1|'''<u>Return to Sample Exam</u>''']] |
