1. ## 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.

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

Cheers

2. Hello, jordanrs!

We need this theorem:

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

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

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