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| | <span class="exam">(c) Evaluate <math style="vertical-align: -14px">\lim _{x\rightarrow -3^+} \frac{x}{x^2-9} </math> | | <span class="exam">(c) Evaluate <math style="vertical-align: -14px">\lim _{x\rightarrow -3^+} \frac{x}{x^2-9} </math> |
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| | + | [[009A Sample Midterm 1, Problem 1 Solution|'''<u>Solution</u>''']] |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| + | [[009A Sample Midterm 1, Problem 1 Detailed Solution|'''<u>Detailed Solution</u>''']] |
| − | !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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| − | | <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.''' Recall
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| − | |-
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| − | | <math style="vertical-align: -14px">\lim_{x\rightarrow 0} \frac{\sin x}{x}=1</math>
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| − | |}
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| − | '''Solution:'''
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| − |
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| − | '''(a)'''
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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 2} x =2\ne 0,</math>
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| − | |-
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| − | |we have
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| − | |-
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| − | | <math>\begin{array}{rcl}
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| − | \displaystyle{5} & = & \displaystyle{\lim _{x\rightarrow 2} \bigg[\frac{4-g(x)}{x}\bigg]}\\
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| − | &&\\
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| − | & = & \displaystyle{\frac{\displaystyle{\lim_{x\rightarrow 2} (4-g(x))}}{\displaystyle{\lim_{x\rightarrow 2} x}}}\\
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| − | &&\\
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| − | & = & \displaystyle{\frac{\displaystyle{\lim_{x\rightarrow 2} (4-g(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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| − | |If we multiply both sides of the last equation by <math>2,</math> we get
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| − | |-
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| − | | <math>10=\lim_{x\rightarrow 2} (4-g(x)).</math>
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| − | |-
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| − | |Now, using linearity properties of limits, we have
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| − | |-
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| − | | <math>\begin{array}{rcl}
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| − | \displaystyle{10} & = & \displaystyle{\lim_{x\rightarrow 2} 4 -\lim_{x\rightarrow 2}g(x)}\\
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| − | &&\\
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| − | & = & \displaystyle{4-\lim_{x\rightarrow 2} g(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 2} g(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 2} g(x)=-6.</math>
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| − | |}
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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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| − | |-
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| − | |First, we write
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| − | |-
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| − | | <math>\lim_{x\rightarrow 0} \frac{\sin(4x)}{5x}=\lim_{x\rightarrow 0} \frac{4}{5} \bigg(\frac{\sin(4x)}{4x}\bigg).</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 0} \frac{\sin(4x)}{5x}} & = & \displaystyle{\frac{4}{5}\lim_{x\rightarrow 0} \frac{\sin(4x)}{4x}}\\
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| − | &&\\
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| − | & = & \displaystyle{\frac{4}{5}(1)}\\
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| − | &&\\
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| − | & = & \displaystyle{\frac{4}{5}.}
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| − | \end{array}</math>
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| − | |}
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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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| − | |-
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| − | |When we plug in values close to <math style="vertical-align: 0px">-3</math> into <math style="vertical-align: -12px">\frac{x}{x^2-9},</math>
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| − | |-
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| − | |we get a small denominator, which results in a large number.
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| − | |-
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| − | |Thus,
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| − | |-
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| − | | <math>\lim_{x\rightarrow -3^+} \frac{x}{x^2-9}</math>
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| − | |-
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| − | |is either equal to <math style="vertical-align: -1px">\infty</math> or <math style="vertical-align: -1px">-\infty.</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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| − | |To figure out which one, we factor the denominator to get
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| − | |-
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| − | | <math>\lim_{x\rightarrow -3^+} \frac{x}{x^2-9}=\lim_{x\rightarrow -3^+} \frac{x}{(x-3)(x+3)}.</math>
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| − | |-
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| − | |We are taking a right hand limit. So, we are looking at values of <math style="vertical-align: 0px">x</math>
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| − | |-
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| − | |a little bigger than <math style="vertical-align: 0px">-3.</math> (You can imagine values like <math style="vertical-align: 0px">x=-2.9.</math> )
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| − | |-
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| − | |For these values, the numerator will be negative.
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| − | |-
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| − | |Also, for these values, <math style="vertical-align: 0px">x-3</math> will be negative and <math style="vertical-align: -1px">x+3</math> will be positive.
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| − | |-
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| − | |Therefore, the denominator will be negative.
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| − | |-
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| − | |Since both the numerator and denominator will be negative (have the same sign),
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| − | |-
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| − | | <math>\lim_{x\rightarrow -3^+} \frac{x}{x^2-9}=\infty.</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> -6</math>
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| − | | '''(b)''' <math>\frac{4}{5}</math>
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| − | |-
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| − | | '''(c)''' <math>\infty</math>
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| − | |}
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| | [[009A_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']] | | [[009A_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']] |
![{\displaystyle \lim _{x\rightarrow 2}{\bigg [}{\frac {4-g(x)}{x}}{\bigg ]}=5.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/81d23dbe64c4cb46da5b3a09f0267a3cfbe494f1)
