# Finding the limit

• Mar 16th 2009, 03:38 PM
jordanrs
Finding the limit
If anyone can help with the following problem id be greatly appreciated. We have just done l'hopitals rule and other limit problems but cant get my head around this one.

Find the following limit.

$\displaystyle\lim_{x\to\infty} \frac{x^{\frac{-4}{3}}}{sin{\frac{1}{x}}}$

Cheers
• Mar 16th 2009, 04:43 PM
Soroban
Hello, jordanrs!

We need this theorem:

. . $\lim_{u\to0}\frac{\sin u}{u} \;=\;\lim_{u\to0}\frac{u}{\sin u} \;=\;1$

Quote:

Find the following limit: . $\lim_{x\to\infty} \frac{x^{-\frac{4}{3}}}{\sin{\frac{1}{x}}}$
Let: . $u = \frac{1}{x}$
Then we have: . $\lim_{u\to0}\frac{u^{\frac{4}{3}}}{\sin u} \;\;=\;\;\lim_{u\to0}\frac{u^{\frac{1}{3}}\cdot u}{\sin u} \;\;=\;\;\lim_{u\to0}\left(u^{\frac{1}{3}}\cdot\fr ac{u}{\sin u}\right)$

. . . . . . . . . $= \;\;\underbrace{ \left(\lim_{u\to0}u^{\frac{1}{3}}\right)}_{\text{T his is 0}}\cdot\underbrace{\left(\lim_{u\to0}\frac{u}{\si n u}\right)}_{\text{This is 1}} \;\;=\;\;0\cdot1 \;\;=\;\;0$