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> | ||
|} | |} | ||
| Line 30: | Line 35: | ||
!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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| Line 66: | Line 71: | ||
!Final Answer: | !Final Answer: | ||
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| − | |'''(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) |