Difference between revisions of "009A Sample Final 3, Problem 7"

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(Created page with "<span class="exam">Compute <span class="exam">(a)  <math style="vertical-align: -18px">\lim_{x\rightarrow 0} \frac{x}{3-\sqrt{9-x}}</math> <span class="exam">(b)  ...")
 
 
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!Foundations: &nbsp;  
 
!Foundations: &nbsp;  
 
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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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|
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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;
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&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>
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&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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|&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 
|&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
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\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})}}\\
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\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})}}\\
 
&&\\
 
&&\\
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& = & \displaystyle{\lim_{x\rightarrow 0} \frac{x(3+\sqrt{9+x})}{9-(9+x)}}\\
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& = & \displaystyle{\lim_{x\rightarrow 0} \frac{x(3+\sqrt{9-x})}{9-(9-x)}}\\
 
&&\\
 
&&\\
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& = & \displaystyle{\lim_{x\rightarrow 0} \frac{x(3+\sqrt{9+x})}{-x}}\\
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& = & \displaystyle{\lim_{x\rightarrow 0} \frac{x(3+\sqrt{9-x})}{x}}\\
 
&&\\
 
&&\\
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& = & \displaystyle{\lim_{x\rightarrow 0} \frac{3+\sqrt{9+x}}{-1}}\\
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& = & \displaystyle{\lim_{x\rightarrow 0} \frac{3+\sqrt{9-x}}{1}}\\
 
&&\\
 
&&\\
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& = & \displaystyle{ \frac{3+\sqrt{9}}{-1}}\\
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& = & \displaystyle{ \frac{3+\sqrt{9}}{1}}\\
 
&&\\
 
&&\\
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& = & \displaystyle{-\frac{6}{1}}\\
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& = & \displaystyle{\frac{6}{1}}\\
 
&&\\
 
&&\\
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& = & \displaystyle{-6.}
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& = & \displaystyle{6.}
 
\end{array}</math>
 
\end{array}</math>
 
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!Final Answer: &nbsp;  
 
!Final Answer: &nbsp;  
 
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|&nbsp; &nbsp;'''(a)'''&nbsp; &nbsp;<math>-6</math>
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|&nbsp; &nbsp;'''(a)'''&nbsp; &nbsp;<math>6</math>
 
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|-
 
|&nbsp; &nbsp;'''(b)'''&nbsp; &nbsp;<math>1</math>
 
|&nbsp; &nbsp;'''(b)'''&nbsp; &nbsp;<math>1</math>

Latest revision as of 17:14, 20 May 2017

Compute

(a)  

(b)  

(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 -2} \frac{x^2-x-6}{x^3+8}}

Foundations:  
L'Hôpital's Rule, Part 1

        Let  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}f(x)=0}   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 \lim_{x\rightarrow c}g(x)=0,}   where  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 f}   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 g}   are differentiable functions

       on an open interval  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 I}   containing  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 c,}   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 g'(x)\ne 0}   on  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 I}   except possibly at  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 c.}  
       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:  
We begin by noticing that we plug in  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 x=0}   into
        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{x}{3-\sqrt{9-x}},}
we get   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{0}{0}.}
Step 2:  
Now, we multiply the numerator and denominator by the conjugate of the denominator.
Hence, 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{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}}

(b)

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 \pi} \frac{\sin (x)}{\pi-x}} & \overset{L'H}{=} & \displaystyle{\lim_{x\rightarrow \pi}\frac{\cos(x)}{-1}.} \end{array}}

Step 2:  
Now, we plug in  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 x=\pi}   to get
        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 \pi} \frac{\sin (x)}{\pi-x}} & = & \displaystyle{\frac{\cos(\pi)}{-1}}\\ &&\\ & = & \displaystyle{\frac{-1}{-1}}\\ &&\\ & = & \displaystyle{1.} \end{array}}

(c)

Step 1:  
We begin by factoring the numerator and denominator. 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 \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)}.}

So, we can cancel  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 x+2}   in the numerator and denominator. Thus, 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 \lim_{x\rightarrow -2} \frac{x^2-x-6}{x^3+8}\,=\,\lim_{x\rightarrow -2}\frac{x-3}{x^2-2x+4}.}

Step 2:  
Now, we can just plug in  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 x=-2}   to get
        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 -2} \frac{x^2-x-6}{x^3+8}} & = & \displaystyle{\frac{-2-3}{(-2)^2-2(-2)+4}}\\ &&\\ & = & \displaystyle{-\frac{5}{12}.} \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 6}
   (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 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 -\frac{5}{12}}

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