009C Sample Final 1, Problem 1

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Compute

a)

b)

Foundations:  
Review L'Hopital's Rule

Solution:

(a)

Step 1:  
First, we switch to the limit to so that we can use L'Hopital's rule.
So, we have
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\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 (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\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 so that we can use L'Hopital's rule.
So, we have
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\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 (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \lim _{n\rightarrow \infty }{\frac {\ln n}{\ln 3n}}=1} .
Final Answer:  
(a) Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\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}

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