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

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!Step 1:    
 
!Step 1:    
 
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|Let
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&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
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\displaystyle{y} & = & \displaystyle{\lim_{n\rightarrow \infty} \bigg(1+\frac{1}{2n}\bigg)^n.}
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\end{array}</math>
 
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|We then take the natural log of both sides to get
 
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|&nbsp; &nbsp; &nbsp; &nbsp; <math>\ln y = \ln\bigg(\lim_{n\rightarrow \infty} \bigg(1+\frac{1}{2n}\bigg)^n\bigg).</math>
 
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
!Step 2: &nbsp;
 
!Step 2: &nbsp;
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|-
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|We can interchange limits and continuous functions.
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|-
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|Therefore, we have
 
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&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
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\displaystyle{\ln y} & = & \displaystyle{\lim_{n\rightarrow \infty} \ln \bigg(1+\frac{1}{2n}\bigg)^n}\\
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&&\\
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& = & \displaystyle{\lim_{n\rightarrow \infty} n\ln\bigg(1+\frac{1}{2n}\bigg)}\\
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&&\\
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& = & \displaystyle{\lim_{n\rightarrow \infty} \frac{\ln \bigg(1+\frac{1}{2n}\bigg)}{\frac{1}{n}}.}
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\end{array}</math>
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|Now, this limit has the form &nbsp;<math style="vertical-align: -13px">\frac{0}{0}.</math>
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|-
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|Hence, we can use L'Hopital's Rule to calculate this limit.
 
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Revision as of 16:25, 5 March 2017

Which of the following sequences  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)_{n\ge 1}}   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.

(b)

Step 1:  
Step 2:  


Final Answer:  
   (a)
   (b)

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