# Proof that \lim_{x->0}\frac{x^2.sin(\frac{1}{x})}{sin(x)}=0

• Oct 31st 2012, 07:02 AM
Chipset3600
Proof that \lim_{x->0}\frac{x^2.sin(\frac{1}{x})}{sin(x)}=0
Hello MHF, please help me how to proof that $\lim_{x->0}\frac{x^2.sin(\frac{1}{x})}{sin(x)}=0$, the book says that L'Hospital Rule doenst work in this case.

Thanks!
• Oct 31st 2012, 07:56 AM
TheEmptySet
Re: Proof that \lim_{x->0}\frac{x^2.sin(\frac{1}{x})}{sin(x)}=0
Quote:

Originally Posted by Chipset3600
Hello MHF, please help me how to proof that $\lim_{x->0}\frac{x^2.sin(\frac{1}{x})}{sin(x)}=0$, the book says that L'Hospital Rule doenst work in this case.

Thanks!

Note that $-1 \le \sin(\left( \frac{1}{x}\right) \le 1$

Now if we multiply by $\frac{x^2}{\sin(x)}$ we get

$-\frac{x^2}{\sin(x)} \le \frac{x^2\sin\left( \frac{1}{x}\right)}{\sin(x)} \le \frac{x^2}{\sin(x)}$

Note that the limit

$\lim_{x \to 0}\frac{x^2}{\sin(x)}=0$

So by the squeeze theorem the limit is 0
• Oct 31st 2012, 07:57 AM
ebaines
Re: Proof that \lim_{x->0}\frac{x^2.sin(\frac{1}{x})}{sin(x)}=0
I would suggest splitting the fraction in two - we know that the limit of x/sin(x) as x goes to 0 is 1. So that leaves the limit of xsin(1/x). The value of sin(1/x) fluctuates between -1 and 1 as x goes to 0. So the limit of x times that is 0. Hence the answer is 0.
• Oct 31st 2012, 03:58 PM
Chipset3600
Re: Proof that \lim_{x->0}\frac{x^2.sin(\frac{1}{x})}{sin(x)}=0
wow! understood, I would not find this solution so quickly. Thanks guys :)