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

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!Step 1:    
 
!Step 1:    
 
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|We begin by factoring the numerator and denominator. We have
 
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&nbsp; &nbsp; &nbsp; &nbsp;<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>
 
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|So, we can cancel &nbsp;<math style="vertical-align: -2px">x+2</math>&nbsp; in the numerator and denominator. Thus, we have
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&nbsp; &nbsp; &nbsp; &nbsp;<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>
 
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!Step 2: &nbsp;
 
!Step 2: &nbsp;
 
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|-
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|Now, we can just plug in &nbsp;<math style="vertical-align: -1px">x=-2</math>&nbsp; to get
 
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|&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
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\displaystyle{\lim_{x\rightarrow -2} \frac{x^2-x-6}{x^3+8}} & = & \displaystyle{\frac{-2-3}{(-2)^2-2(-2)+4}}\\
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&&\\
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& = & \displaystyle{\frac{-5}{12}.}
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\end{array}</math>
 
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|'''(b)'''
 
|'''(b)'''
 
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|'''(c)'''
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|&nbsp; &nbsp;'''(c)'''&nbsp; &nbsp;<math>\frac{-5}{12}</math>
 
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[[009A_Sample_Final_3|'''<u>Return to Sample Exam</u>''']]
 
[[009A_Sample_Final_3|'''<u>Return to Sample Exam</u>''']]

Revision as of 12:26, 7 March 2017

Compute

(a)  

(b)  

(c)  

Foundations:  
L'Hôpital's Rule
        Suppose that    and    are both zero or both  

        If    is finite or  

        then  


Solution:

(a)

Step 1:  
Step 2:  

(b)

Step 1:  
Step 2:  

(c)

Step 1:  
We begin by factoring the numerator and denominator. We have

       

So, we can cancel    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)
(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 \frac{-5}{12}}

Return to Sample Exam