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!
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
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.