Difference between revisions of "009C Sample Final 3, Problem 1"

From Grad Wiki
Jump to navigation Jump to search
Line 57: Line 57:
 
|-
 
|-
 
|Hence, we can use L'Hopital's Rule to calculate this limit.
 
|Hence, we can use L'Hopital's Rule to calculate this limit.
 +
|}
 +
 +
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 +
!Step 3:  
 +
|-
 +
|Now, we have
 
|-
 
|-
 
|
 
|
 +
&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 +
\displaystyle{\ln y } & = & \displaystyle{\lim_{n\rightarrow \infty} \frac{\ln \bigg(1+\frac{1}{2n}\bigg)}{\frac{1}{n}}}\\
 +
&&\\
 +
& = & \displaystyle{\lim_{x\rightarrow \infty} \frac{\ln \bigg(1+\frac{1}{2x}\bigg)}{\frac{1}{x}}}\\
 +
&&\\
 +
& \overset{L'H}{=} & \displaystyle{\lim_{x\rightarrow \infty} \frac{\frac{2x}{2x+1}\big(\frac{-1}{2x^2}\big)}{\big(-\frac{1}{x^2}\big)}}\\
 +
&&\\
 +
& = & \displaystyle{\lim_{x\rightarrow \infty} \frac{1}{2}\bigg(\frac{2x}{2x+1}\bigg)}\\
 +
&&\\
 +
& = & \displaystyle{\frac{1}{2}.}
 +
\end{array}</math>
 +
|}
 +
 +
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 +
!Step 4: &nbsp;
 +
|-
 +
|Since &nbsp;<math style="vertical-align: -13px">\ln y= \frac{1}{2},</math>&nbsp; we know
 +
|-
 +
|&nbsp; &nbsp; &nbsp; &nbsp; <math>y=e^{\frac{1}{2}}.</math>
 
|}
 
|}
  
Line 85: Line 110:
 
!Final Answer: &nbsp;  
 
!Final Answer: &nbsp;  
 
|-
 
|-
|&nbsp;&nbsp; '''(a)'''  
+
|&nbsp; &nbsp; '''(a)''' &nbsp; &nbsp;<math>e^{\frac{1}{2}}</math>
 
|-
 
|-
|&nbsp;&nbsp; '''(b)'''  
+
|&nbsp; &nbsp; '''(b)''' &nbsp; &nbsp;
 
|}
 
|}
 
[[009C_Sample_Final_3|'''<u>Return to Sample Exam</u>''']]
 
[[009C_Sample_Final_3|'''<u>Return to Sample Exam</u>''']]

Revision as of 16:33, 5 March 2017

Which of the following sequences    converges? Which diverges? Give reasons for your answers!

(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=\bigg(1+\frac{1}{2n}\bigg)^n}

(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=\cos(n\pi)\bigg(\frac{1+n}{n}\bigg)^n}

Foundations:  
L'Hôpital'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:  
Let

        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{y} & = & \displaystyle{\lim_{n\rightarrow \infty} \bigg(1+\frac{1}{2n}\bigg)^n.} \end{array}}

We then take the natural log of both sides to 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 \ln y = \ln\bigg(\lim_{n\rightarrow \infty} \bigg(1+\frac{1}{2n}\bigg)^n\bigg).}
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(1+\frac{1}{2n}\bigg)^n}\\ &&\\ & = & \displaystyle{\lim_{n\rightarrow \infty} n\ln\bigg(1+\frac{1}{2n}\bigg)}\\ &&\\ & = & \displaystyle{\lim_{n\rightarrow \infty} \frac{\ln \bigg(1+\frac{1}{2n}\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(1+\frac{1}{2n}\bigg)}{\frac{1}{n}}}\\ &&\\ & = & \displaystyle{\lim_{x\rightarrow \infty} \frac{\ln \bigg(1+\frac{1}{2x}\bigg)}{\frac{1}{x}}}\\ &&\\ & \overset{L'H}{=} & \displaystyle{\lim_{x\rightarrow \infty} \frac{\frac{2x}{2x+1}\big(\frac{-1}{2x^2}\big)}{\big(-\frac{1}{x^2}\big)}}\\ &&\\ & = & \displaystyle{\lim_{x\rightarrow \infty} \frac{1}{2}\bigg(\frac{2x}{2x+1}\bigg)}\\ &&\\ & = & \displaystyle{\frac{1}{2}.} \end{array}}

Step 4:  
Since  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \ln y={\frac {1}{2}},}   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^{\frac{1}{2}}.}

(b)

Step 1:  
Step 2:  


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 e^{\frac{1}{2}}}
    (b)    

Return to Sample Exam