Difference between revisions of "009C Sample Final 2, Problem 1"
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!Step 1: | !Step 1: | ||
|- | |- | ||
− | | | + | |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> |
|} | |} | ||
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!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> | ||
|} | |} | ||
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!Final Answer: | !Final Answer: | ||
|- | |- | ||
− | | '''(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 |
Suppose that and are both zero or both |
If is finite or |
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: |
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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) |