Difference between revisions of "009C Sample Final 1, Problem 1"
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!Foundations: | !Foundations: | ||
|- | |- | ||
− | | | + | |Review L'Hopital's Rule |
|} | |} | ||
Line 18: | Line 18: | ||
!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}}\\ | ||
+ | &&\\ | ||
+ | & = & \displaystyle{\frac{-2}{5}} | ||
+ | \end{array}</math> | ||
|} | |} | ||
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!Step 2: | !Step 2: | ||
|- | |- | ||
− | | | + | |Hence, we have |
|- | |- | ||
− | | | + | |<math>\lim_{n\rightarrow \infty} \frac{3-2n^2}{5n^2+n+1}=\frac{-2}{5}</math> |
|- | |- | ||
| | | | ||
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!Final Answer: | !Final Answer: | ||
|- | |- | ||
− | |'''(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: |
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First, we switch to the limit to so that we can use L'Hopital's rule. |
So, we have |
|
Step 2: |
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Hence, we have |
(b)
Step 1: |
---|
Step 2: |
---|
Step 3: |
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Final Answer: |
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(a) |
(b) |