009A Sample Midterm 3, Problem 1 - Revision history

MathAdmin at 22:23, 9 November 2017

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MathAdmin: Created page with " Find the following limits: (a) If $\lim _{x\rightarrow 3} \bigg(\frac{f(x)}{2x}+1\bigg)=2,$
2017-04-09T18:54:51Z

Created page with " Find the following limits: (a) If <math style="vertical-align: -16px">\lim _{x\rightarrow 3} \bigg(\frac{f(x)}{2x}+1\bigg)=2,</ma..."

New page
 Find the following limits:

(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>

(b) Find  <math style="vertical-align: -19px">\lim _{x\rightarrow 0} \frac{\tan(4x)}{\sin(6x)}. </math>

(c) Evaluate  <math style="vertical-align: -16px">\lim _{x\rightarrow \infty} \frac{-2x^3-2x+3}{3x^3+3x^2-3}. </math>

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
|-
|'''1.''' If  <math style="vertical-align: -13px">\lim_{x\rightarrow a} g(x)\neq 0,</math>  we have
|-
|        <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>
|-
|'''2.''' Recall
|-
|       <math style="vertical-align: -15px">\lim_{x\rightarrow 0} \frac{\sin x}{x}=1</math>
|}

'''Solution:'''

'''(a)'''
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|First, we have
|-
|        <math>\begin{array}{rcl}
\displaystyle{2} & = & \displaystyle{\lim_{x\rightarrow 3} \bigg(\frac{f(x)}{2x}+1\bigg)}\\
&&\\
& = & \displaystyle{\lim_{x\rightarrow 3} \frac{f(x)}{2x}+\lim_{x\rightarrow 3} 1}\\
&&\\
& = & \displaystyle{\lim_{x\rightarrow 3} \frac{f(x)}{2x}+1.}
\end{array}</math>
|-
|Therefore,
|-
|        <math>\lim_{x\rightarrow 3} \frac{f(x)}{2x}=1.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Since  <math style="vertical-align: -13px">\lim_{x\rightarrow 3} 2x=6\ne 0,</math>  we have
|-
|
        <math>\begin{array}{rcl}
\displaystyle{1} & = & \displaystyle{\lim_{x\rightarrow 3} \frac{f(x)}{2x}}\\
&&\\
& = & \displaystyle{\frac{\lim_{x\rightarrow 3} f(x)}{\lim_{x\rightarrow} 2x}}\\
&&\\
& = & \displaystyle{\frac{\lim_{x\rightarrow 3} f(x)}{6}.}
\end{array}</math>
|-
|Multiplying both sides by  <math style="vertical-align: -5px">6,</math>  we get
|-
|        <math>\lim_{x\rightarrow 3} f(x)=6.</math>
|}

'''(b)'''
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|First, we write
|-
|        <math>\begin{array}{rcl}
\displaystyle{\lim_{x\rightarrow 0} \frac{\tan(4x)}{\sin(6x)}} & = & \displaystyle{\lim_{x\rightarrow 0} \frac{\sin(4x)}{\cos(4x)} \frac{1}{\sin(6x)}}\\
&&\\
& = & \displaystyle{\lim_{x\rightarrow 0} \frac{4}{6} \frac{\sin(4x)}{4x}\frac{6x}{\sin(6x)}\frac{1}{\cos(4x)}}\\
&&\\
& = & \displaystyle{\frac{4}{6}\lim_{x\rightarrow 0} \frac{\sin(4x)}{4x}\frac{6x}{\sin(6x)}\frac{1}{\cos(4x)}.}
\end{array}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Now, we have
|-
|
        <math>\begin{array}{rcl}
\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)}}\\
&&\\
& = & \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)}\\
&&\\
& = & \displaystyle{\frac{4}{6} (1)(1)(1)}\\
&&\\
& = & \displaystyle{\frac{2}{3}.}
\end{array}</math>
|}

'''(c)'''
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|First, we have
|-
|        <math>\begin{array}{rcl}
\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})}}\\
&&\\
& = & \displaystyle{\lim_{x\rightarrow 0} \frac{-2-\frac{2}{x^2}+\frac{3}{x^3}}{3+\frac{3}{x}-\frac{3}{x^3}}}.
\end{array}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Now, we use the properties of limits to get
|-
|
        <math>\begin{array}{rcl}
\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}}}\\
&&\\
& = & \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})}}\\
&&\\
& = & \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}}} \\
&&\\
& = & \displaystyle{\frac{-2+0+0}{3+0+0}}\\
&&\\
& = & \displaystyle{-\frac{2}{3}.}
\end{array}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|    '''(a)'''     <math>6</math>
|-
|    '''(b)'''     <math>\frac{2}{3}</math>
|-
|    '''(c)'''     <math>-\frac{2}{3}</math>
|}
[[009A_Sample_Midterm_3|'''Return to Sample Exam''']]

@@ Line 8: / Line 8: @@
 <hr>
-[[File:9ASM3P1.jpg|600px|thumb|center]]
+[[009A Sample Midterm 3, Problem 1 Detailed Solution|'''<u>Detailed Solution</u>''']]
 [[009A_Sample_Midterm_3|'''<u>Return to Sample Exam</u>''']]

@@ Line 7: / Line 7: @@
 <span class="exam">(c) Evaluate &nbsp;<math style="vertical-align: -16px">\lim _{x\rightarrow \infty} \frac{-2x^3-2x+3}{3x^3+3x^2-3}. </math>
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
+[[009A Sample Midterm 3, Problem 1 Detailed Solution|'''<u>Detailed Solution with Background Information</u>''']]
 [[009A_Sample_Midterm_3|'''<u>Return to Sample Exam</u>''']]

@@ Line 102: / Line 102: @@
 \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})}}\\
 &&\\
-& = & \displaystyle{\lim_{x\rightarrow 0} \frac{-2-\frac{2}{x^2}+\frac{3}{x^3}}{3+\frac{3}{x}-\frac{3}{x^3}}}.
+& = & \displaystyle{\lim_{x\rightarrow \infty} \frac{-2-\frac{2}{x^2}+\frac{3}{x^3}}{3+\frac{3}{x}-\frac{3}{x^3}}}.
 \end{array}</math>
 |}
@@ Line 115: / Line 115: @@
 \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}}}\\
 &&\\
-& = & \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})}}\\
+& = & \displaystyle{\frac{\displaystyle{\lim_{x\rightarrow \infty} \bigg(-2-\frac{2}{x^2}+\frac{3}{x^3}\bigg)}}{\displaystyle{\lim_{x\rightarrow \infty} \bigg(3+\frac{3}{x}-\frac{3}{x^3}\bigg)}}}\\
 &&\\
-& = & \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}}} \\
+& = & \displaystyle{\frac{\displaystyle{\lim_{x\rightarrow \infty} -2 +\lim_{x\rightarrow \infty} \frac{2}{x^2} +\lim_{x\rightarrow \infty} \frac{3}{x^3}}}{\displaystyle{\lim_{x\rightarrow \infty} 3+\lim_{x\rightarrow \infty} \frac{3}{x}-\lim_{x\rightarrow \infty}\frac{3}{x^3}}}} \\
 &&\\
 & = & \displaystyle{\frac{-2+0+0}{3+0+0}}\\

@@ Line 51: / Line 51: @@
 \displaystyle{1} & = & \displaystyle{\lim_{x\rightarrow 3} \frac{f(x)}{2x}}\\
 &&\\
-& = & \displaystyle{\frac{\lim_{x\rightarrow 3} f(x)}{\lim_{x\rightarrow} 2x}}\\
+& = & \displaystyle{\frac{\displaystyle{\lim_{x\rightarrow 3} f(x)}}{\displaystyle{\lim_{x\rightarrow 3} 2x}}}\\
 &&\\
-& = & \displaystyle{\frac{\lim_{x\rightarrow 3} f(x)}{6}.}
+& = & \displaystyle{\frac{\displaystyle{\lim_{x\rightarrow 3} f(x)}}{6}.}
 \end{array}</math>
 |-