009C Sample Final 1, Problem 1 Detailed Solution

Compute

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

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

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Background Information:
L'Hopital's Rule helps calculate limits that have indeterminate forms
${\frac {0}{0}}$  or  ${\frac {\pm \infty }{\infty }}.$

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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