Difference between revisions of "009C Sample Final 2, Problem 1"
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!Step 1: | !Step 1: | ||
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| − | | | + | |First, we notice that <math>\lim_{n\rightarrow \infty} \frac{\ln(n)}{\ln(n+1)}</math> has the form <math>\frac{\infty}{\infty}.</math> |
|- | |- | ||
| − | | | + | |So, we can use L'Hopital's Rule. To begin, we write |
|- | |- | ||
| − | | | + | | <math>\lim_{n\rightarrow \infty}\frac{\ln(n)}{\ln(n+1)}=\lim_{x\rightarrow \infty} \frac{\ln(x)}{\ln(x+1)}.</math> |
|} | |} | ||
| Line 38: | Line 38: | ||
!Step 2: | !Step 2: | ||
|- | |- | ||
| − | | | + | |Now, using L'Hopital's rule, we get |
|- | |- | ||
| − | | | + | | <math>\begin{array}{rcl} |
| + | \displaystyle{\lim_{n\rightarrow \infty}\frac{\ln(n)}{\ln(n+1)}} & \overset{L'H}{=} & \displaystyle{\lim_{x\rightarrow \infty} \frac{\frac{1}{x}}{\frac{1}{x+1}}}\\ | ||
| + | &&\\ | ||
| + | & = & \displaystyle{\lim_{x\rightarrow \infty} \frac{x+1}{x}}\\ | ||
| + | &&\\ | ||
| + | & = & \displaystyle{\lim_{x\rightarrow \infty} 1+\frac{1}{x}}\\ | ||
| + | &&\\ | ||
| + | & = & \displaystyle{1.} | ||
| + | \end{array}</math> | ||
|} | |} | ||
| Line 67: | Line 75: | ||
!Final Answer: | !Final Answer: | ||
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| − | | '''(a)''' | + | | '''(a)''' <math>1</math> |
|- | |- | ||
| '''(b)''' | | '''(b)''' | ||
|} | |} | ||
[[009C_Sample_Final_2|'''<u>Return to Sample Exam</u>''']] | [[009C_Sample_Final_2|'''<u>Return to Sample Exam</u>''']] | ||
Revision as of 18:32, 4 March 2017
Test if the following sequences converge or diverge. Also find the limit of each convergent sequence.
(a)
(b)
| Foundations: |
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| L'Hopital's Rule |
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Suppose that and are both zero or both |
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If is finite or |
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then |
Solution:
(a)
| Step 1: |
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| First, we notice that has the form |
| So, we can use L'Hopital's Rule. To begin, we write |
| Step 2: |
|---|
| Now, using L'Hopital's rule, we get |
(b)
| Step 1: |
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| Step 2: |
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| Final Answer: |
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| (a) |
| (b) |