009C Sample Final 1, Problem 1

Compute

(a)   $\lim _{n\rightarrow \infty }{\frac {3-2n^{2}}{5n^{2}+n+1}}$

(b)   $\lim _{n\rightarrow \infty }{\frac {\ln n}{\ln 3n}}$

Foundations:
L'Hopital's Rule
Suppose that $\lim _{x\rightarrow \infty }f(x)$   and $\lim _{x\rightarrow \infty }g(x)$   are both zero or both   $\pm \infty .$
If $\lim _{x\rightarrow \infty }{\frac {f'(x)}{g'(x)}}$   is finite or   $\pm \infty ,$
then $\lim _{x\rightarrow \infty }{\frac {f(x)}{g(x)}}=\lim _{x\rightarrow \infty }{\frac {f'(x)}{g'(x)}}.$

Solution:

(a)

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

Step 2:
Hence, we have
$\lim _{n\rightarrow \infty }{\frac {3-2n^{2}}{5n^{2}+n+1}}={\frac {-2}{5}}.$

(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