# 009C Sample Final 1, Problem 1

Compute

a) ${\displaystyle \lim _{n\rightarrow \infty }{\frac {3-2n^{2}}{5n^{2}+n+1}}}$
b) ${\displaystyle \lim _{n\rightarrow \infty }{\frac {\ln n}{\ln 3n}}}$
Foundations:
Recall:
L'Hopital's Rule
Suppose that ${\displaystyle \lim _{x\rightarrow \infty }f(x)}$ and ${\displaystyle \lim _{x\rightarrow \infty }g(x)}$ are both zero or both ${\displaystyle \pm \infty .}$
If ${\displaystyle \lim _{x\rightarrow \infty }{\frac {f'(x)}{g'(x)}}}$ is finite or ${\displaystyle \pm \infty ,}$
then ${\displaystyle \lim _{x\rightarrow \infty }{\frac {f(x)}{g(x)}}=\lim _{x\rightarrow \infty }{\frac {f'(x)}{g'(x)}}.}$

Solution:

(a)

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

(b)

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