Difference between revisions of "009A Sample Final 3, Problem 1"

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(Created page with "<span class="exam">Find each of the following limits if it exists. If you think the limit does not exist provide a reason. <span class="exam">(a)  <math style="vertical-...")
 
 
(One intermediate revision by the same user not shown)
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\displaystyle{-2} & = & \displaystyle{\lim _{x\rightarrow 8} \bigg[\frac{xf(x)}{3}\bigg]}\\
 
\displaystyle{-2} & = & \displaystyle{\lim _{x\rightarrow 8} \bigg[\frac{xf(x)}{3}\bigg]}\\
 
&&\\
 
&&\\
& = & \displaystyle{\frac{\lim_{x\rightarrow 8} xf(x)}{\lim_{x\rightarrow 8} 3}}\\
+
& = & \displaystyle{\frac{\displaystyle{\lim_{x\rightarrow 8} xf(x)}}{\displaystyle{\lim_{x\rightarrow 8} 3}}}\\
 
&&\\
 
&&\\
& = & \displaystyle{\frac{\lim_{x\rightarrow 8} xf(x)}{3}.}
+
& = & \displaystyle{\frac{\displaystyle{\lim_{x\rightarrow 8} xf(x)}}{3}.}
 
\end{array}</math>
 
\end{array}</math>
 
|-
 
|-
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|-
 
|-
 
|&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 
|&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
\displaystyle{\lim_{x\rightarrow -\infty} \frac{\sqrt{9x^6-x}}{3x^3+4x}} & = & \displaystyle{\lim_{x\rightarrow -\infty} \frac{\sqrt{9x^6-x}}{3x^3+4x}\frac{\big(\frac{1}{x^3}\big)}{\big(\frac{1}{x^3}\big)}}\\
+
\displaystyle{\lim_{x\rightarrow -\infty} \frac{\sqrt{9x^6-x}}{3x^3+4x}} & = & \displaystyle{\lim_{x\rightarrow -\infty} \frac{\sqrt{9x^6(1-\frac{1}{9x^5})}}{3x^3+4x}}\\
 
&&\\
 
&&\\
& = & \displaystyle{\lim_{x\rightarrow -\infty} \frac{\sqrt{9-\frac{1}{x^5}}}{3+\frac{4}{x^2}}.}
+
& = & \displaystyle{\lim_{x\rightarrow -\infty} \frac{3|x^3|\sqrt{(1-\frac{1}{9x^5})}}{3x^3+4x}}\\
 +
&&\\
 +
& = & \displaystyle{\lim_{x\rightarrow -\infty} \frac{3(-x^3)\sqrt{(1-\frac{1}{9x^5})}}{3x^3(1+\frac{4}{3x^2})}}
 
\end{array}</math>
 
\end{array}</math>
 
|}
 
|}
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|-
 
|-
 
|&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 
|&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
\displaystyle{\lim_{x\rightarrow -\infty} \frac{\sqrt{9x^6-x}}{3x^3+4x}} & = & \displaystyle{\frac{\lim_{x\rightarrow -\infty} \sqrt{9-\frac{1}{x^5}}}{\lim_{x\rightarrow -\infty}3+\frac{4}{x^2}}}\\
+
\displaystyle{\lim_{x\rightarrow -\infty} \frac{\sqrt{9x^6-x}}{3x^3+4x}} & = & \displaystyle{\lim_{x\rightarrow -\infty} \frac{3(-x^3)\sqrt{(1-\frac{1}{9x^5})}}{3x^3(1+\frac{4}{3x^2})}}\\
 +
&&\\
 +
& = & \displaystyle{\lim_{x\rightarrow -\infty} \frac{-\sqrt{(1-\frac{1}{9x^5})}}{1+\frac{4}{3x^2}}}\\
 
&&\\
 
&&\\
& = & \displaystyle{\frac{\sqrt{9}}{3}}\\
+
& = & \displaystyle{\frac{-\sqrt{1}}{1}}\\
 
&&\\
 
&&\\
& = & \displaystyle{1.}
+
& = & \displaystyle{-1.}\\
 
\end{array}</math>
 
\end{array}</math>
 
|}
 
|}
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|&nbsp; &nbsp;'''(b)'''&nbsp; &nbsp; <math>-\frac{3}{4}</math>
 
|&nbsp; &nbsp;'''(b)'''&nbsp; &nbsp; <math>-\frac{3}{4}</math>
 
|-
 
|-
|&nbsp; &nbsp;'''(c)'''&nbsp; &nbsp; <math>1</math>
+
|&nbsp; &nbsp;'''(c)'''&nbsp; &nbsp; <math>-1</math>
 
|}
 
|}
 
[[009A_Sample_Final_3|'''<u>Return to Sample Exam</u>''']]
 
[[009A_Sample_Final_3|'''<u>Return to Sample Exam</u>''']]

Latest revision as of 09:07, 4 December 2017

Find each of the following limits if it exists. If you think the limit does not exist provide a reason.

(a)  

(b)    given that  

(c)  


Foundations:  
1. If    we have
       
2.  


Solution:

(a)

Step 1:  
We begin by noticing that we plug in    into
       
we get  
Step 2:  
Now, we multiply the numerator and denominator by the conjugate of the denominator.
Hence, we have
       

(b)

Step 1:  
Since  
we have
       
Step 2:  
If we multiply both sides of the last equation by    we get
       
Now, using properties of limits, we have
       
Step 3:  
Solving for    in the last equation,
we get

       

(c)

Step 1:  
First, we write
       
Step 2:  
Now, we have
       


Final Answer:  
   (a)   
   (b)   
   (c)   

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