Difference between revisions of "009A Sample Final 2, Problem 8"

From Grad Wiki
Jump to navigation Jump to search
 
(One intermediate revision by the same user not shown)
Line 6: Line 6:
  
 
<span class="exam">(c) &nbsp;<math style="vertical-align: -15px">\lim_{x\rightarrow 1} \frac{x^3-1}{x^{10}-1}</math>
 
<span class="exam">(c) &nbsp;<math style="vertical-align: -15px">\lim_{x\rightarrow 1} \frac{x^3-1}{x^{10}-1}</math>
 +
<hr>
 +
[[009A Sample Final 2, Problem 8 Solution|'''<u>Solution</u>''']]
  
−
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
−
!Foundations: &nbsp;
 
−
|-
 
−
|'''L'Hôpital's Rule'''
 
−
|-
 
−
|&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; If &nbsp;<math style="vertical-align: -19px">\lim_{x\rightarrow \infty} \frac{f'(x)}{g'(x)}</math>&nbsp; is finite or &nbsp;<math style="vertical-align: -4px">\pm \infty ,</math>
 
−
|-
 
−
|
 
−
&nbsp; &nbsp; &nbsp; &nbsp; then &nbsp;<math style="vertical-align: -19px">\lim_{x\rightarrow \infty} \frac{f(x)}{g(x)}\,=\,\lim_{x\rightarrow \infty} \frac{f'(x)}{g'(x)}.</math>
 
−
|}
 
  
 +
[[009A Sample Final 2, Problem 8 Detailed Solution|'''<u>Detailed Solution</u>''']]
  
−
'''Solution:'''
 
  
−
'''(a)'''
 
−
 
−
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
−
!Step 1: &nbsp;
 
−
|-
 
−
|First, we have
 
−
|-
 
−
|&nbsp; &nbsp; &nbsp; &nbsp; <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: &nbsp;
 
−
|-
 
−
|Now, we have
 
−
|-
 
−
|&nbsp; &nbsp; &nbsp; &nbsp; <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: &nbsp;
 
−
|-
 
−
|First, we write
 
−
|-
 
−
|&nbsp; &nbsp; &nbsp; &nbsp; <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: &nbsp;
 
−
|-
 
−
|Now, we have
 
−
|-
 
−
|&nbsp; &nbsp; &nbsp; &nbsp; <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
 
−
|-
 
−
|&nbsp; &nbsp; &nbsp; &nbsp; <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,
 
−
|-
 
−
|&nbsp; &nbsp; &nbsp; &nbsp;<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: &nbsp;
 
−
|-
 
−
|We proceed using L'Hôpital's Rule. So, we have
 
−
|-
 
−
|
 
−
&nbsp; &nbsp; &nbsp; &nbsp;<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: &nbsp;
 
−
|-
 
−
|Now, we have
 
−
|-
 
−
|&nbsp; &nbsp; &nbsp; &nbsp;<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: &nbsp;
 
−
|-
 
−
|&nbsp; &nbsp;'''(a)'''&nbsp; &nbsp; <math>\infty</math>
 
−
|-
 
−
|&nbsp; &nbsp;'''(b)'''&nbsp; &nbsp; <math>\text{DNE}</math>
 
−
|-
 
−
|&nbsp; &nbsp;'''(c)'''&nbsp; &nbsp; <math>\frac{3}{10}</math>
 
−
|}
 
 
[[009A_Sample_Final_2|'''<u>Return to Sample Exam</u>''']]
 
[[009A_Sample_Final_2|'''<u>Return to Sample Exam</u>''']]

Latest revision as of 09:23, 1 December 2017

Compute

(a)  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{x\rightarrow \infty} \frac{x^{-1}+x}{1+\sqrt{1+x}}}

(b)  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{x\rightarrow 0} \frac{\sin x}{\cos x-1}}

(c)  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{x\rightarrow 1} \frac{x^3-1}{x^{10}-1}}

Solution


Detailed Solution


Return to Sample Exam

Retrieved from "https://gradwiki.math.ucr.edu/index.php?title=009A_Sample_Final_2,_Problem_8&oldid=4247"