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

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|Hence, we have
 
|Hence, we have
 
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|-
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|<math>\lim_{n\rightarrow \infty} \frac{3-2n^2}{5n^2+n+1}=\frac{-2}{5}</math>
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|<math>\lim_{n\rightarrow \infty} \frac{3-2n^2}{5n^2+n+1}=\frac{-2}{5}</math>.
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|-
 
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|}
 
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!Step 1: &nbsp;  
 
!Step 1: &nbsp;  
 
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|-
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|Again, we switch to the limit to <math>x</math> so that we can use L'Hopital's rule.
 
|-
 
|-
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|
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|So, we have
 
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|-
 
|
 
|
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::<math>\begin{array}{rcl}
 +
\displaystyle{\lim_{x \rightarrow \infty}\frac{\ln x}{\ln(3x)}} & \overset{l'H}{=} & \displaystyle{\lim_{x \rightarrow \infty}\frac{\frac{1}{x}}{\frac{3}{3x}}}\\
 +
&&\\
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& = & \displaystyle{\lim_{x \rightarrow \infty} 1}\\
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&&\\
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& = & 1
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\end{array}</math>
 
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!Step 2: &nbsp;
 
!Step 2: &nbsp;
 
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|-
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|
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|Hence, we have
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|}
 
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
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!Step 3: &nbsp;
 
 
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|<math>\lim_{n\rightarrow \infty} \frac{\ln n}{\ln 3n}=1</math>.
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|'''(a)''' <math>\frac{-2}{5}</math>
 
|'''(a)''' <math>\frac{-2}{5}</math>
 
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|-
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|'''(b)'''  
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|'''(b)''' <math>1</math>
 
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[[009C_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]
 
[[009C_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]

Revision as of 17:34, 8 February 2016

Compute

a)

b)

Foundations:  
Review L'Hopital's Rule

Solution:

(a)

Step 1:  
First, we switch to the limit to 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 x} so that we can use L'Hopital's rule.
So, 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{\lim_{x \rightarrow \infty}\frac{3-2x^2}{5x^2 + x +1}} & \overset{l'H}{=} & \displaystyle{\lim_{x \rightarrow \infty}\frac{-4x}{10x+1}}\\ &&\\ & \overset{l'H}{=} & \displaystyle{\frac{-4}{10}}\\ &&\\ & = & \displaystyle{\frac{-2}{5}} \end{array}}
Step 2:  
Hence, 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 \lim_{n\rightarrow \infty} \frac{3-2n^2}{5n^2+n+1}=\frac{-2}{5}} .

(b)

Step 1:  
Again, we switch to the limit to 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 x} so that we can use L'Hopital's rule.
So, 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{\lim_{x \rightarrow \infty}\frac{\ln x}{\ln(3x)}} & \overset{l'H}{=} & \displaystyle{\lim_{x \rightarrow \infty}\frac{\frac{1}{x}}{\frac{3}{3x}}}\\ &&\\ & = & \displaystyle{\lim_{x \rightarrow \infty} 1}\\ &&\\ & = & 1 \end{array}}
Step 2:  
Hence, 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 \lim_{n\rightarrow \infty} \frac{\ln n}{\ln 3n}=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 \frac{-2}{5}}
(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 1}

Return to Sample Exam

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