Difference between revisions of "009C Sample Final 1, Problem 1"

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!Foundations:    
 
!Foundations:    
 
|-
 
|-
|
+
|Review L'Hopital's Rule
 
|}
 
|}
  
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!Step 1:    
 
!Step 1:    
 
|-
 
|-
|
+
|First, we switch to the limit to <math>x</math> so that we can use L'Hopital's rule.
 
|-
 
|-
|
+
|So, we have
|-
 
|
 
 
|-
 
|-
 
|
 
|
 +
::<math>\begin{array}{rcl}
 +
\displaystyle{\lim_{x \rightarrow \infty}\frac{3-2x^2}{5x^2 + x +1}} & \overset{l'H}{=} & \displaystyle{\lim_{x \rightarrow \infty}\frac{-4x}{10x+1}}\\
 +
&&\\
 +
& \overset{l'H}{=} & \displaystyle{\frac{-4}{10}}\\
 +
&&\\
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& = & \displaystyle{\frac{-2}{5}}
 +
\end{array}</math>
 
|}
 
|}
  
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!Step 2: &nbsp;
 
!Step 2: &nbsp;
 
|-
 
|-
|
+
|Hence, we have
 
|-
 
|-
|
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|<math>\lim_{n\rightarrow \infty} \frac{3-2n^2}{5n^2+n+1}=\frac{-2}{5}</math>
 
|-
 
|-
 
|
 
|
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!Final Answer: &nbsp;  
 
!Final Answer: &nbsp;  
 
|-
 
|-
|'''(a)'''
+
|'''(a)''' <math>\frac{-2}{5}</math>
 
|-
 
|-
 
|'''(b)'''  
 
|'''(b)'''  
 
|}
 
|}
 
[[009C_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]
 
[[009C_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]

Revision as of 18:29, 8 February 2016

Compute

a)

b)

Foundations:  
Review L'Hopital's Rule

Solution:

(a)

Step 1:  
First, we switch to the limit to so that we can use L'Hopital's rule.
So, we have
Step 2:  
Hence, we have

(b)

Step 1:  
Step 2:  
Step 3:  
Final Answer:  
(a)
(b)

Return to Sample Exam