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> |
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| Line 38: | Line 38: | ||
!Step 2: | !Step 2: | ||
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| − | | | + | |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: | ||
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| − | | '''(a)''' | + | | '''(a)''' <math>1</math> |
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| '''(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) Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a_n=\frac{\ln(n)}{\ln(n+1)}}
(b) Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a_n=\bigg(\frac{n}{n+1}\bigg)^n}
| Foundations: |
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| L'Hopital's Rule |
|
Suppose that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{x\rightarrow \infty} f(x)} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{x\rightarrow \infty} g(x)} are both zero or both Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \pm \infty .} |
|
If Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{x\rightarrow \infty} \frac{f'(x)}{g'(x)}} is finite or Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \pm \infty ,} |
|
then Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{x\rightarrow \infty} \frac{f(x)}{g(x)}=\lim_{x\rightarrow \infty} \frac{f'(x)}{g'(x)}.} |
Solution:
(a)
| Step 1: |
|---|
| First, we notice that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{n\rightarrow \infty} \frac{\ln(n)}{\ln(n+1)}} has the form Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{\infty}{\infty}.} |
| So, we can use L'Hopital's Rule. To begin, we write |
| Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{n\rightarrow \infty}\frac{\ln(n)}{\ln(n+1)}=\lim_{x\rightarrow \infty} \frac{\ln(x)}{\ln(x+1)}.} |
| Step 2: |
|---|
| Now, using L'Hopital's rule, we get |
| Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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}} |
(b)
| Step 1: |
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| Step 2: |
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| Final Answer: |
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| (a) Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 1} |
| (b) |