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 $x$ so that we can use L'Hopital's rule.
So, we have
${\begin{array}{rcl}\displaystyle {\lim _{x\rightarrow \infty }{\frac {\ln x}{\ln(3x)}}}&{\overset {l'H}{=}}&\displaystyle {\lim _{x\rightarrow \infty }{\frac {\frac {1}{x}}{\frac {3}{3x}}}}\\&&\\&=&\displaystyle {\lim _{x\rightarrow \infty }1}\\&&\\&=&1.\end{array}}$

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

Final Answer:
(a)     ${\frac {-2}{5}}$
(b)     $1$

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