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

**Chipset3600** · October 31st 2012, 07:02 AM

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!

**TheEmptySet** · October 31st 2012, 07:56 AM

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

**ebaines** · October 31st 2012, 07:57 AM

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.

**Chipset3600** · October 31st 2012, 03:58 PM

wow! understood, I would not find this solution so quickly. Thanks guys

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