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

Thanks!

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- Oct 31st 2012, 07:02 AMChipset3600Proof that \lim_{x->0}\frac{x^2.sin(\frac{1}{x})}{sin(x)}=0
Hello MHF, please help me how to proof that $\displaystyle \lim_{x->0}\frac{x^2.sin(\frac{1}{x})}{sin(x)}=0$, the book says that L'Hospital Rule doenst work in this case.

Thanks! - Oct 31st 2012, 07:56 AMTheEmptySetRe: Proof that \lim_{x->0}\frac{x^2.sin(\frac{1}{x})}{sin(x)}=0
Note that $\displaystyle -1 \le \sin(\left( \frac{1}{x}\right) \le 1$

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

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

Note that the limit

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

So by the squeeze theorem the limit is 0 - Oct 31st 2012, 07:57 AMebainesRe: 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 PMChipset3600Re: 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 :)