Difference between revisions of "009C Sample Final 2, Problem 1"
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| <math>\begin{array}{rcl} | | <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_{n\rightarrow \infty}\frac{\ln(n)}{\ln(n+1)}} & \overset{L'H}{=} & \displaystyle{\lim_{x\rightarrow \infty} \frac{\big(\frac{1}{x}\big)}{\big(\frac{1}{x+1}\big)}}\\ |
&&\\ | &&\\ | ||
& = & \displaystyle{\lim_{x\rightarrow \infty} \frac{x+1}{x}}\\ | & = & \displaystyle{\lim_{x\rightarrow \infty} \frac{x+1}{x}}\\ | ||
Revision as of 12:15, 18 March 2017
Test if the following sequences converge or diverge. Also find the limit of each convergent sequence.
(a)
(b)
| Foundations: |
|---|
| L'Hopital's Rule |
|
Suppose that and are both zero or both |
|
If is finite or |
|
then |
Solution:
(a)
| Step 1: |
|---|
| 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 |
| Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\begin{array}{rcl}\displaystyle {\lim _{n\rightarrow \infty }{\frac {\ln(n)}{\ln(n+1)}}}&{\overset {L'H}{=}}&\displaystyle {\lim _{x\rightarrow \infty }{\frac {{\big (}{\frac {1}{x}}{\big )}}{{\big (}{\frac {1}{x+1}}{\big )}}}}\\&&\\&=&\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: |
|---|
| Let
|
| We then take the natural log of both sides to get |
| Step 2: |
|---|
| We can interchange limits and continuous functions. |
| Therefore, we have |
|
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{\ln y} & = & \displaystyle{\lim_{n\rightarrow \infty} \ln \bigg(\frac{n}{n+1}\bigg)^n}\\ &&\\ & = & \displaystyle{\lim_{n\rightarrow \infty} n\ln\bigg(\frac{n}{n+1}\bigg)}\\ &&\\ & = & \displaystyle{\lim_{n\rightarrow \infty} \frac{\ln \bigg(\frac{n}{n+1}\bigg)}{\frac{1}{n}}.} \end{array}} |
| Now, this limit 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{0}{0}.} |
| Hence, we can use L'Hopital's Rule to calculate this limit. |
| Step 3: |
|---|
| Now, we have |
|
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{\ln y } & = & \displaystyle{\lim_{n\rightarrow \infty} \frac{\ln \bigg(\frac{n}{n+1}\bigg)}{\frac{1}{n}}}\\ &&\\ & = & \displaystyle{\lim_{x\rightarrow \infty} \frac{\ln \bigg(\frac{x}{x+1}\bigg)}{\frac{1}{x}}}\\ &&\\ & \overset{L'H}{=} & \displaystyle{\lim_{x\rightarrow \infty} \frac{\frac{x+1}{x}\frac{1}{(x+1)^2}}{\big(-\frac{1}{x^2}\big)}}\\ &&\\ & = & \displaystyle{\lim_{x\rightarrow \infty} \frac{-x^2}{x(x+1)}}\\ &&\\ & = & \displaystyle{-1.} \end{array}} |
| Step 4: |
|---|
| Since 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 \ln y= -1,} we know |
| 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 y=e^{-1}.} |
| Final Answer: |
|---|
| (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) 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 e^{-1}} |