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

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!Foundations:    
 
!Foundations:    
 
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|'''L'Hôpital's Rule'''  
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|'''L'Hôpital's Rule, Part 1'''  
 
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|&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>
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&nbsp; &nbsp; &nbsp; &nbsp; Let &nbsp;<math style="vertical-align: -12px">\lim_{x\rightarrow c}f(x)=0</math>&nbsp; and &nbsp;<math style="vertical-align: -12px">\lim_{x\rightarrow c}g(x)=0,</math>&nbsp; where &nbsp;<math style="vertical-align: -5px">f</math>&nbsp; and &nbsp;<math style="vertical-align: -5px">g</math>&nbsp; are differentiable functions
 
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|&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; 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>
 
 
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|&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>
&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>
 
 
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Revision as of 16:39, 20 May 2017

Compute

(a)  

(b)  

(c)  

Foundations:  
L'Hôpital's Rule, Part 1

        Let    and    where    and    are differentiable functions

       on an open interval    containing    and    on    except possibly at   
       Then,   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 c} \frac{f(x)}{g(x)}=\lim_{x\rightarrow c} \frac{f'(x)}{g'(x)}.}


Solution:

(a)

Step 1:  
First, we have
        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 \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}}
Step 2:  
Now, we have
        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 \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}}

(b)

Step 1:  
First, we write
        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 \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}}
Step 2:  
Now, we have
        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 \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}}
and
        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 \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}}
Therefore,
       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}=\text{DNE}.}

(c)

Step 1:  
We proceed using L'Hôpital's Rule. So, we have

       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 \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}}

Step 2:  
Now, we have
       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 \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}}


Final Answer:  
   (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 \infty}
   (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 \text{DNE}}
   (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 \frac{3}{10}}

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