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| − | <span class="exam">Find the following limits: | + | <span class="exam"> Find the following limits: |
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| − | <span class="exam">(a) If <math>\lim _{x\rightarrow 3} \bigg(\frac{f(x)}{2x}+1\bigg)=2,</math> find <math>\lim _{x\rightarrow 3} f(x).</math> | + | <span class="exam">(a) If <math style="vertical-align: -16px">\lim _{x\rightarrow 3} \bigg(\frac{f(x)}{2x}+1\bigg)=2,</math> find <math style="vertical-align: -13px">\lim _{x\rightarrow 3} f(x).</math> |
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| − | <span class="exam">(b) Find <math>\lim _{x\rightarrow 0} \frac{\tan(4x)}{\sin(6x)}. </math> | + | <span class="exam">(b) Find <math style="vertical-align: -19px">\lim _{x\rightarrow 0} \frac{\tan(4x)}{\sin(6x)}. </math> |
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| − | <span class="exam">(c) Evaluate <math>\lim _{x\rightarrow \infty} \frac{-2x^3-2x+3}{3x^3+3x^2-3}. </math> | + | <span class="exam">(c) Evaluate <math style="vertical-align: -16px">\lim _{x\rightarrow \infty} \frac{-2x^3-2x+3}{3x^3+3x^2-3}. </math> |
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| | + | <hr> |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| + | [[009A Sample Midterm 3, Problem 1 Detailed Solution|'''<u>Detailed Solution with Background Information</u>''']] |
| − | !Foundations:
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| − | |- | |
| − | |'''1.''' If <math>\lim_{x\rightarrow a} g(x)\neq 0</math>, we have
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| − | |-
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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.''' <math>\lim_{x\rightarrow 0} \frac{\sin x}{x}=1</math>
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| − | |}
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| | + | [[File:9ASM3P1.jpg|600px|thumb|center]] |
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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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| − | |First, we have
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| − | |-
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| − | | <math>\begin{array}{rcl}
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| − | \displaystyle{2} & = & \displaystyle{\lim_{x\rightarrow 3} \bigg(\frac{f(x)}{2x}+1\bigg)}\\
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| − | &&\\
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| − | & = & \displaystyle{\lim_{x\rightarrow 3} \frac{f(x)}{2x}+\lim_{x\rightarrow 3} 1}\\
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| − | &&\\
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| − | & = & \displaystyle{\lim_{x\rightarrow 3} \frac{f(x)}{2x}+1.}
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| − | \end{array}</math>
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| − | |-
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| − | |Therefore,
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| − | |-
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| − | | <math>\lim_{x\rightarrow 3} \frac{f(x)}{2x}=1.</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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| − | |Since <math>\lim_{x\rightarrow 3} 2x=6\ne 0,</math> we have
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| − | |-
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| − | |
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| − | <math>\begin{array}{rcl}
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| − | \displaystyle{1} & = & \displaystyle{\lim_{x\rightarrow 3} \frac{f(x)}{2x}}\\
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| − | &&\\
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| − | & = & \displaystyle{\frac{\lim_{x\rightarrow 3} f(x)}{\lim_{x\rightarrow} 2x}}\\
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| − | &&\\
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| − | & = & \displaystyle{\frac{\lim_{x\rightarrow 3} f(x)}{6}.}
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| − | \end{array}</math>
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| − | |-
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| − | |Multiplying both sides by <math>6,</math> we get
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| − | |-
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| − | | <math>\lim_{x\rightarrow 3} f(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>\begin{array}{rcl}
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| − | \displaystyle{\lim_{x\rightarrow 0} \frac{\tan(4x)}{\sin(6x)}} & = & \displaystyle{\lim_{x\rightarrow 0} \frac{\sin(4x)}{\cos(4x)} \frac{1}{\sin(6x)}}\\
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| − | &&\\
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| − | & = & \displaystyle{\lim_{x\rightarrow 0} \frac{4}{6} \frac{\sin(4x)}{4x}\frac{6x}{\sin(6x)}\frac{1}{\cos(4x)}}\\
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| − | &&\\
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| − | & = & \displaystyle{\frac{4}{6}\lim_{x\rightarrow 0} \frac{\sin(4x)}{4x}\frac{6x}{\sin(6x)}\frac{1}{\cos(4x)}.}
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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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| − | |Now, 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{\tan(4x)}{\sin(6x)}} & = & \displaystyle{\frac{4}{6}\lim_{x\rightarrow 0} \frac{\sin(4x)}{4x}\frac{6x}{\sin(6x)}\frac{1}{\cos(4x)}}\\
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| − | &&\\
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| − | & = & \displaystyle{\frac{4}{6}\bigg(\lim_{x\rightarrow 0} \frac{\sin(4x)}{4x}\bigg)\bigg(\lim_{x\rightarrow 0} \frac{6x}{\sin(6x)}\bigg)\bigg(\lim_{x\rightarrow 0} \frac{1}{\cos(4x)}\bigg)}\\
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| − | &&\\
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| − | & = & \displaystyle{\frac{4}{6} (1)(1)(1)}\\
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| − | &&\\
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| − | & = & \displaystyle{\frac{2}{3}.}
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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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| − | |First, we have
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| − | |-
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| − | | <math>\begin{array}{rcl}
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| − | \displaystyle{\lim _{x\rightarrow \infty} \frac{-2x^3-2x+3}{3x^3+3x^2-3}} & = & \displaystyle{\lim _{x\rightarrow \infty} \frac{(-2x^3-2x+3)}{(3x^3+3x^2-3)} \frac{(\frac{1}{x^3})}{(\frac{1}{x^3})}}\\
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| − | &&\\
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| − | & = & \displaystyle{\lim_{x\rightarrow 0} \frac{-2-\frac{2}{x^2}+\frac{3}{x^3}}{3+\frac{3}{x}-\frac{3}{x^3}}}.
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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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| − | |Now, we use the properties of limits to get
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| − | |-
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| − | |
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| − | <math>\begin{array}{rcl}
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| − | \displaystyle{\lim _{x\rightarrow \infty} \frac{-2x^3-2x+3}{3x^3+3x^2-3}} & = & \displaystyle{\lim_{x\rightarrow \infty} \frac{-2-\frac{2}{x^2}+\frac{3}{x^3}}{3+\frac{3}{x}-\frac{3}{x^3}}}\\
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| − | &&\\
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| − | & = & \displaystyle{\frac{\lim_{x\rightarrow \infty} (-2-\frac{2}{x^2}+\frac{3}{x^3})}{\lim_{x\rightarrow \infty} (3+\frac{3}{x}-\frac{3}{x^3})}}\\
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| − | &&\\
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| − | & = & \displaystyle{\frac{\lim_{x\rightarrow \infty} -2 +\lim_{x\rightarrow \infty} \frac{2}{x^2} +\lim_{x\rightarrow \infty} \frac{3}{x^3}}{\lim_{x\rightarrow \infty} 3+\lim_{x\rightarrow \infty} \frac{3}{x}-\lim_{x\rightarrow \infty}\frac{3}{x^3}}} \\
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| − | &&\\
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| − | & = & \displaystyle{\frac{-2+0+0}{3+0+0}}\\
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| − | &&\\
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| − | & = & \displaystyle{-\frac{2}{3}.}
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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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| − | !Final Answer:
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| − | |-
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| − | | '''(a)''' <math>\lim_{x\rightarrow 3} f(x)=6</math>
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| − | |-
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| − | | '''(b)''' <math>\frac{2}{3}</math>
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| − | |-
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| − | | '''(c)''' <math>-\frac{2}{3}</math>
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| − | |}
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| | [[009A_Sample_Midterm_3|'''<u>Return to Sample Exam</u>''']] | | [[009A_Sample_Midterm_3|'''<u>Return to Sample Exam</u>''']] |
