MathAdmin at 01:14, 21 May 2017

2017-05-21T01:14:56Z

MathAdmin: Created page with "Compute (a) $\lim_{x\rightarrow 0} \frac{x}{3-\sqrt{9-x}}$ (b) ..."

2017-04-10T16:31:27Z

Created page with "Compute (a) <math style="vertical-align: -18px">\lim_{x\rightarrow 0} \frac{x}{3-\sqrt{9-x}}</math> (b) ..."

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Compute

(a)  <math style="vertical-align: -18px">\lim_{x\rightarrow 0} \frac{x}{3-\sqrt{9-x}}</math>

(b)  <math style="vertical-align: -16px">\lim_{x\rightarrow \pi} \frac{\sin x}{\pi-x}</math>

(c)  <math style="vertical-align: -16px">\lim_{x\rightarrow -2} \frac{x^2-x-6}{x^3+8}</math>

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
|-
|'''L'Hôpital's Rule'''
|-
|        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>
|-
|
        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>
|-
|
        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>
|}

'''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{x}{3-\sqrt{9-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{x}{3-\sqrt{9-x}}} & = & \displaystyle{\lim_{x\rightarrow 0} \frac{x}{3-\sqrt{9-x}}\frac{(3+\sqrt{9+x})}{(3+\sqrt{9+x})}}\\
&&\\
& = & \displaystyle{\lim_{x\rightarrow 0} \frac{x(3+\sqrt{9+x})}{9-(9+x)}}\\
&&\\
& = & \displaystyle{\lim_{x\rightarrow 0} \frac{x(3+\sqrt{9+x})}{-x}}\\
&&\\
& = & \displaystyle{\lim_{x\rightarrow 0} \frac{3+\sqrt{9+x}}{-1}}\\
&&\\
& = & \displaystyle{ \frac{3+\sqrt{9}}{-1}}\\
&&\\
& = & \displaystyle{-\frac{6}{1}}\\
&&\\
& = & \displaystyle{-6.}
\end{array}</math>
|-
|
|}

'''(b)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|We proceed using L'Hôpital's Rule. So, we have
|-
|
       <math>\begin{array}{rcl}
\displaystyle{\lim_{x\rightarrow \pi} \frac{\sin (x)}{\pi-x}} & \overset{L'H}{=} & \displaystyle{\lim_{x\rightarrow \pi}\frac{\cos(x)}{-1}.}
\end{array}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Now, we plug in  <math style="vertical-align: 0px">x=\pi</math>  to get
|-
|        <math>\begin{array}{rcl}
\displaystyle{\lim_{x\rightarrow \pi} \frac{\sin (x)}{\pi-x}} & = & \displaystyle{\frac{\cos(\pi)}{-1}}\\
&&\\
& = & \displaystyle{\frac{-1}{-1}}\\
&&\\
& = & \displaystyle{1.}
\end{array}</math>
|}

'''(c)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|We begin by factoring the numerator and denominator. We have
|-
|
       <math>\lim_{x\rightarrow -2} \frac{x^2-x-6}{x^3+8}\,=\,\lim_{x\rightarrow -2}\frac{(x+2)(x-3)}{(x+2)(x^2-2x+4)}.</math>
|-
|So, we can cancel  <math style="vertical-align: -2px">x+2</math>  in the numerator and denominator. Thus, we have
|-
|
       <math>\lim_{x\rightarrow -2} \frac{x^2-x-6}{x^3+8}\,=\,\lim_{x\rightarrow -2}\frac{x-3}{x^2-2x+4}.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Now, we can just plug in  <math style="vertical-align: -1px">x=-2</math>  to get
|-
|        <math>\begin{array}{rcl}
\displaystyle{\lim_{x\rightarrow -2} \frac{x^2-x-6}{x^3+8}} & = & \displaystyle{\frac{-2-3}{(-2)^2-2(-2)+4}}\\
&&\\
& = & \displaystyle{-\frac{5}{12}.}
\end{array}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|   '''(a)'''   <math>-6</math>
|-
|   '''(b)'''   <math>1</math>
|-
|   '''(c)'''   <math>-\frac{5}{12}</math>
|}
[[009A_Sample_Final_3|'''Return to Sample Exam''']]

@@ Line 10: / Line 10: @@
 !Foundations: &nbsp;
 |-
-|'''L'Hôpital's Rule'''
+|'''L'Hôpital's Rule, Part 1'''
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp; Suppose that &nbsp;<math style="vertical-align: -11px">\lim_{x\rightarrow \infty} f(x)</math>&nbsp; and &nbsp;<math style="vertical-align: -11px">\lim_{x\rightarrow \infty} g(x)</math>&nbsp; are both zero or both &nbsp;<math style="vertical-align: -1px">\pm \infty .</math>
 |-
+|&nbsp; &nbsp; &nbsp; &nbsp;on an open interval &nbsp;<math style="vertical-align: 0px">I</math>&nbsp; containing &nbsp;<math style="vertical-align: -5px">c,</math>&nbsp; and &nbsp;<math style="vertical-align: -5px">g'(x)\ne 0</math>&nbsp; on &nbsp;<math style="vertical-align: 0px">I</math>&nbsp; except possibly at &nbsp;<math style="vertical-align: 0px">c.</math>&nbsp;
 |-
+|&nbsp; &nbsp; &nbsp; &nbsp;Then, &nbsp; <math style="vertical-align: -18px">\lim_{x\rightarrow c} \frac{f(x)}{g(x)}=\lim_{x\rightarrow c} \frac{f'(x)}{g'(x)}.</math>
 |}
@@ Line 44: / Line 43: @@
 |-
 |&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
-\displaystyle{\lim_{x\rightarrow 0} \frac{x}{3-\sqrt{9-x}}} & = & \displaystyle{\lim_{x\rightarrow 0} \frac{x}{3-\sqrt{9-x}}\frac{(3+\sqrt{9+x})}{(3+\sqrt{9+x})}}\\
+\displaystyle{\lim_{x\rightarrow 0} \frac{x}{3-\sqrt{9-x}}} & = & \displaystyle{\lim_{x\rightarrow 0} \frac{x}{3-\sqrt{9-x}}\frac{(3+\sqrt{9-x})}{(3+\sqrt{9-x})}}\\
 &&\\
-& = & \displaystyle{\lim_{x\rightarrow 0} \frac{x(3+\sqrt{9+x})}{9-(9+x)}}\\
+& = & \displaystyle{\lim_{x\rightarrow 0} \frac{x(3+\sqrt{9-x})}{9-(9-x)}}\\
 &&\\
-& = & \displaystyle{\lim_{x\rightarrow 0} \frac{x(3+\sqrt{9+x})}{-x}}\\
+& = & \displaystyle{\lim_{x\rightarrow 0} \frac{x(3+\sqrt{9-x})}{x}}\\
 &&\\
-& = & \displaystyle{\lim_{x\rightarrow 0} \frac{3+\sqrt{9+x}}{-1}}\\
+& = & \displaystyle{\lim_{x\rightarrow 0} \frac{3+\sqrt{9-x}}{1}}\\
 &&\\
-& = & \displaystyle{ \frac{3+\sqrt{9}}{-1}}\\
+& = & \displaystyle{ \frac{3+\sqrt{9}}{1}}\\
 &&\\
-& = & \displaystyle{-\frac{6}{1}}\\
+& = & \displaystyle{\frac{6}{1}}\\
 &&\\
-& = & \displaystyle{-6.}
+& = & \displaystyle{6.}
 \end{array}</math>
 |-
@@ Line 121: / Line 120: @@
 !Final Answer: &nbsp;
 |-
-|&nbsp; &nbsp;'''(a)'''&nbsp; &nbsp;<math>-6</math>
+|&nbsp; &nbsp;'''(a)'''&nbsp; &nbsp;<math>6</math>
 |-
 |&nbsp; &nbsp;'''(b)'''&nbsp; &nbsp;<math>1</math>

009A Sample Final 3, Problem 7 - Revision history

MathAdmin at 01:14, 21 May 2017

MathAdmin: Created page with "Compute (a) \lim_{x\rightarrow 0} \frac{x}{3-\sqrt{9-x}} (b) ..."

MathAdmin: Created page with "Compute (a) $\lim_{x\rightarrow 0} \frac{x}{3-\sqrt{9-x}}$ (b) ..."