MathAdmin at 01:12, 21 May 2017

2017-05-21T01:12:29Z

MathAdmin: Created page with "Compute (a) $\lim_{x\rightarrow \infty} \frac{x^{-1}+x}{1+\sqrt{1+x}}$
2017-04-10T16:25:36Z

Created page with "Compute (a) <math style="vertical-align: -18px">\lim_{x\rightarrow \infty} \frac{x^{-1}+x}{1+\sqrt{1+x}}</math> Compute

(a)  <math style="vertical-align: -18px">\lim_{x\rightarrow \infty} \frac{x^{-1}+x}{1+\sqrt{1+x}}</math>

(b)  <math style="vertical-align: -15px">\lim_{x\rightarrow 0} \frac{\sin x}{\cos x-1}</math>

(c)  <math style="vertical-align: -15px">\lim_{x\rightarrow 1} \frac{x^3-1}{x^{10}-1}</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:  
|-
|First, we have
|-
|        <math>\begin{array}{rcl}
\displaystyle{\lim_{x\rightarrow \infty} \frac{x^{-1}+x}{1+\sqrt{1+x}}} & = & \displaystyle{\lim_{x\rightarrow \infty}\frac{\frac{1}{x}+x}{1+\sqrt{1+x}}}\\
&&\\
& = & \displaystyle{\lim_{x\rightarrow \infty}\frac{\frac{1}{x}+x}{1+\sqrt{1+x}} \frac{\big(\frac{1}{\sqrt{x}}\big)}{\big(\frac{1}{\sqrt{x}}\big)}}\\
&&\\
& = & \displaystyle{\lim_{x\rightarrow \infty} \frac{\frac{1}{x^{3/2}}+\sqrt{x}}{\frac{1}{\sqrt{x}}+\sqrt{\frac{1}{x}+1}}.}
\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{x^{-1}+x}{1+\sqrt{1+x}}} & = & \displaystyle{\lim_{x\rightarrow \infty} \frac{\frac{1}{x^{3/2}}+\sqrt{x}}{\frac{1}{\sqrt{x}}+\sqrt{\frac{1}{x}+1}}}\\
&&\\
& = & \displaystyle{\frac{\lim_{x\rightarrow \infty}\big(\frac{1}{x^{3/2}}+\sqrt{x}\big)}{\lim_{x\rightarrow \infty}\big(\frac{1}{\sqrt{x}}+\sqrt{\frac{1}{x}+1}\big)}}\\
&&\\
& = & \displaystyle{\frac{\lim_{x\rightarrow \infty}\frac{1}{x^{3/2}}+\lim_{x\rightarrow \infty}\sqrt{x}}{\lim_{x\rightarrow \infty}\frac{1}{\sqrt{x}}+\lim_{x\rightarrow \infty}\sqrt{\frac{1}{x}+1}}}\\
&&\\
& = & \displaystyle{\frac{0+\lim_{x\rightarrow \infty}\sqrt{x}}{0+1}}\\
&&\\
& = & \displaystyle{\infty.}
\end{array}</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{\sin x}{\cos x-1}} & = & \displaystyle{\lim_{x\rightarrow 0} \frac{\sin x}{\cos x-1}\frac{(\cos x+1)}{(\cos x+1)}}\\
&&\\
& = & \displaystyle{\lim_{x\rightarrow 0} \frac{\sin x (\cos x+1)}{\cos^2x-1}}\\
&&\\
& = & \displaystyle{\lim_{x\rightarrow 0} \frac{\sin x(\cos x+1)}{-\sin^2 x}}\\
&&\\
& = & \displaystyle{\lim_{x\rightarrow 0} \frac{\cos x+1}{-\sin x}.}
\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{\sin x}{\cos x-1}} & = & \displaystyle{\lim_{x\rightarrow 0^+} \frac{\cos x+1}{-\sin x}}\\
&&\\
& = & \displaystyle{-\infty}
\end{array}</math>
|-
|and
|-
|        <math>\begin{array}{rcl}
\displaystyle{\lim_{x\rightarrow 0^-} \frac{\sin x}{\cos x-1}} & = & \displaystyle{\lim_{x\rightarrow 0^-} \frac{\cos x+1}{-\sin x}}\\
&&\\
& = & \displaystyle{\infty.}
\end{array}</math>
|-
|Therefore,
|-
|       <math>\lim_{x\rightarrow 0} \frac{\sin x}{\cos x-1}=\text{DNE}.</math>
|}

'''(c)'''

{| 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 1} \frac{x^3-1}{x^{10}-1}} & \overset{L'H}{=} & \displaystyle{\lim_{x\rightarrow 1}\frac{3x^2}{10x^9}.}
\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 1} \frac{x^3-1}{x^{10}-1}} & \overset{L'H}{=} & \displaystyle{\lim_{x\rightarrow 1}\frac{3x^2}{10x^9}}\\
&&\\
& = & \displaystyle{\frac{3(1)^2}{10(1)^9}}\\
&&\\
& = & \displaystyle{\frac{3}{10}.}
\end{array}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|   '''(a)'''    <math>\infty</math>
|-
|   '''(b)'''    <math>\text{DNE}</math>
|-
|   '''(c)'''    <math>\frac{3}{10}</math>
|}
[[009A_Sample_Final_2|'''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>
 |}