009C Sample Final 1, Problem 1 Detailed Solution

Compute

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

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

---

Background Information:
L'Hopital's Rule helps calculate limits that have indeterminate forms
${\frac {0}{0}}$  or  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\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
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  $x$  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
$\lim _{n\rightarrow \infty }{\frac {\ln n}{\ln 3n}}=1.$

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)     $1$

Return to Sample Exam