009A Sample Final 3, Problem 1 - Revision history

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MathAdmin at 01:14, 21 May 2017

2017-05-21T01:14:03Z

MathAdmin: Created page with "Find each of the following limits if it exists. If you think the limit does not exist provide a reason. (a)
2017-04-10T16:27:56Z

Created page with "Find each of the following limits if it exists. If you think the limit does not exist provide a reason. (a) <math style="vertical-..."

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Find each of the following limits if it exists. If you think the limit does not exist provide a reason.

(a)  <math style="vertical-align: -14px">\lim_{x\rightarrow 0} \frac{\sin(5x)}{1-\sqrt{1-x}}</math>

(b)  <math style="vertical-align: -12px">\lim_{x\rightarrow 8} f(x),</math>  given that  <math style="vertical-align: -14px">\lim_{x\rightarrow 8}\frac{xf(x)}{3}=-2</math>

(c)  <math style="vertical-align: -14px">\lim_{x\rightarrow -\infty} \frac{\sqrt{9x^6-x}}{3x^3+4x}</math>

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
|-
| '''1.''' If  <math style="vertical-align: -12px">\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.'''  <math style="vertical-align: -14px">\lim_{x\rightarrow 0} \frac{\sin x}{x}=1</math>
|}

'''Solution:'''

'''(a)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|We begin by noticing that we plug in  <math style="vertical-align: 0px">x=0</math>  into
|-
|        <math>\frac{\sin(5x)}{1-\sqrt{1-x}},</math>
|-
|we get   <math style="vertical-align: -12px">\frac{0}{0}.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Now, we multiply the numerator and denominator by the conjugate of the denominator.
|-
|Hence, we have
|-
|        <math>\begin{array}{rcl}
\displaystyle{\lim_{x\rightarrow 0} \frac{\sin(5x)}{1-\sqrt{1-x}}} & = & \displaystyle{\lim_{x\rightarrow 0} \frac{\sin(5x)}{1-\sqrt{1-x}} \bigg(\frac{1+\sqrt{1-x}}{1+\sqrt{1-x}}\bigg)}\\
&&\\
& = & \displaystyle{\lim_{x\rightarrow 0} \frac{\sin(5x)(1+\sqrt{1-x})}{x}}\\
&&\\
& = & \displaystyle{\lim_{x\rightarrow 0} \frac{\sin(5x)}{x}(1+\sqrt{1-x})}\\
&&\\
& = & \displaystyle{\bigg(\lim_{x\rightarrow 0} \frac{\sin(5x)}{x}\bigg) \lim_{x\rightarrow 0}(1+\sqrt{1-x})}\\
&&\\
& = & \displaystyle{\bigg(5\lim_{x\rightarrow 0} \frac{\sin(5x)}{5x}\bigg) (2)}\\
&&\\
& = & \displaystyle{5(1)(2)}\\
&&\\
& = & \displaystyle{10.}
\end{array}</math>
|}

'''(b)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|Since  <math style="vertical-align: -12px">\lim_{x\rightarrow 8} 3 =3\ne 0,</math>
|-
|we have
|-
|        <math>\begin{array}{rcl}
\displaystyle{-2} & = & \displaystyle{\lim _{x\rightarrow 8} \bigg[\frac{xf(x)}{3}\bigg]}\\
&&\\
& = & \displaystyle{\frac{\lim_{x\rightarrow 8} xf(x)}{\lim_{x\rightarrow 8} 3}}\\
&&\\
& = & \displaystyle{\frac{\lim_{x\rightarrow 8} xf(x)}{3}.}
\end{array}</math>
|-
|
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|If we multiply both sides of the last equation by  <math>3,</math>  we get
|-
|        <math>-6=\lim_{x\rightarrow 8} xf(x).</math>
|-
|Now, using properties of limits, we have
|-
|        <math>\begin{array}{rcl}
\displaystyle{-6} & = & \displaystyle{\bigg(\lim_{x\rightarrow 8} x\bigg)\bigg(\lim_{x\rightarrow 8}f(x)\bigg)}\\
&&\\
& = & \displaystyle{8\lim_{x\rightarrow 8} f(x).}\\
\end{array}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 3:  
|-
|Solving for  <math style="vertical-align: -12px">\lim_{x\rightarrow 8} f(x)</math>  in the last equation,
|-
|we get
|-
|
        <math> \lim_{x\rightarrow 8} f(x)=-\frac{3}{4}.</math>
|}

'''(c)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|First, we write
|-
|        <math>\begin{array}{rcl}
\displaystyle{\lim_{x\rightarrow -\infty} \frac{\sqrt{9x^6-x}}{3x^3+4x}} & = & \displaystyle{\lim_{x\rightarrow -\infty} \frac{\sqrt{9x^6-x}}{3x^3+4x}\frac{\big(\frac{1}{x^3}\big)}{\big(\frac{1}{x^3}\big)}}\\
&&\\
& = & \displaystyle{\lim_{x\rightarrow -\infty} \frac{\sqrt{9-\frac{1}{x^5}}}{3+\frac{4}{x^2}}.}
\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 -\infty} \frac{\sqrt{9x^6-x}}{3x^3+4x}} & = & \displaystyle{\frac{\lim_{x\rightarrow -\infty} \sqrt{9-\frac{1}{x^5}}}{\lim_{x\rightarrow -\infty}3+\frac{4}{x^2}}}\\
&&\\
& = & \displaystyle{\frac{\sqrt{9}}{3}}\\
&&\\
& = & \displaystyle{1.}
\end{array}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|   '''(a)'''    <math>10</math>
|-
|   '''(b)'''    <math>-\frac{3}{4}</math>
|-
|   '''(c)'''    <math>1</math>
|}
[[009A_Sample_Final_3|'''Return to Sample Exam''']]

@@ Line 112: / Line 112: @@
 |-
 |&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
-\displaystyle{\lim_{x\rightarrow -\infty} \frac{\sqrt{9x^6-x}}{3x^3+4x}} & = & \displaystyle{\lim_{x\rightarrow -\infty} \frac{\sqrt{9x^6-x}}{3x^3+4x}\frac{\big(\frac{1}{x^3}\big)}{\big(\frac{1}{x^3}\big)}}\\
+\displaystyle{\lim_{x\rightarrow -\infty} \frac{\sqrt{9x^6-x}}{3x^3+4x}} & = & \displaystyle{\lim_{x\rightarrow -\infty} \frac{\sqrt{9x^6(1-\frac{1}{9x^5})}}{3x^3+4x}}\\
 &&\\
-& = & \displaystyle{\lim_{x\rightarrow -\infty} \frac{\sqrt{9-\frac{1}{x^5}}}{3+\frac{4}{x^2}}.}
+& = & \displaystyle{\lim_{x\rightarrow -\infty} \frac{3|x^3|\sqrt{(1-\frac{1}{9x^5})}}{3x^3+4x}}\\
 \end{array}</math>
 |}
@@ Line 124: / Line 126: @@
 |-
 |&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
-\displaystyle{\lim_{x\rightarrow -\infty} \frac{\sqrt{9x^6-x}}{3x^3+4x}} & = & \displaystyle{\frac{\lim_{x\rightarrow -\infty} \sqrt{9-\frac{1}{x^5}}}{\lim_{x\rightarrow -\infty}3+\frac{4}{x^2}}}\\
+\displaystyle{\lim_{x\rightarrow -\infty} \frac{\sqrt{9x^6-x}}{3x^3+4x}} & = & \displaystyle{\lim_{x\rightarrow -\infty} \frac{3(-x^3)\sqrt{(1-\frac{1}{9x^5})}}{3x^3(1+\frac{4}{3x^2})}}\\
 &&\\
-& = & \displaystyle{\frac{\sqrt{9}}{3}}\\
+& = & \displaystyle{\frac{-\sqrt{1}}{1}}\\
 &&\\
-& = & \displaystyle{1.}
+& = & \displaystyle{-1.}\\
 \end{array}</math>
 |}
@@ Line 140: / Line 144: @@
 |&nbsp; &nbsp;'''(b)'''&nbsp; &nbsp; <math>-\frac{3}{4}</math>
 |-
-|&nbsp; &nbsp;'''(c)'''&nbsp; &nbsp; <math>1</math>
+|&nbsp; &nbsp;'''(c)'''&nbsp; &nbsp; <math>-1</math>
 |}
 [[009A_Sample_Final_3|'''<u>Return to Sample Exam</u>''']]

@@ Line 69: / Line 69: @@
 \displaystyle{-2} & = & \displaystyle{\lim _{x\rightarrow 8} \bigg[\frac{xf(x)}{3}\bigg]}\\
 &&\\
-& = & \displaystyle{\frac{\lim_{x\rightarrow 8} xf(x)}{\lim_{x\rightarrow 8} 3}}\\
+& = & \displaystyle{\frac{\displaystyle{\lim_{x\rightarrow 8} xf(x)}}{\displaystyle{\lim_{x\rightarrow 8} 3}}}\\
 &&\\
-& = & \displaystyle{\frac{\lim_{x\rightarrow 8} xf(x)}{3}.}
+& = & \displaystyle{\frac{\displaystyle{\lim_{x\rightarrow 8} xf(x)}}{3}.}
 \end{array}</math>
 |-