What is the value of $\lim_{x\to0}\frac{x^2\sin\left(\frac{1}{x}\right)}{\sin x}$? So this is the question in my text book $$\lim_{x\to0}\frac{x^2\sin\left(\frac{1}{x}\right)}{\sin x}$$
here what i have done
It can be rearranged as $$\lim_{x\to0}\left(\frac{x}{\sin x}\right)x\sin\left(\frac{1}{x}\right)$$$$\Rightarrow\lim_{x\to0}(1)x\sin\left(\frac{1}{x}\right)$$$$\Rightarrow\lim_{x\to0}\frac{\sin\left(\frac{1}{x}\right)}{\frac{1}{x}}$$$$\Rightarrow1$$
This is what i got. Have done something wrong cause in my text-book the answer is $0$ 
Here the solution in my text book $$\lim_{x\to0}\left(\frac{x}{\sin x}\right)x\sin\left(\frac{1}{x}\right)$$$$\Rightarrow\lim_{x\to0}(1)x\sin\left(\frac{1}{x}\right)$$$$\Rightarrow0$$
as$$\left|x\sin\frac{1}{x}\right|\le|x|$$
can anyone help me what is wrong 
Thanks
Akash

In your approach, you claim that $\lim_{x \rightarrow 0}\dfrac{\sin\frac{1}{x}}{\frac{1}{x}}=1$, but in fact the limit $\dfrac{\sin t}{t} \rightarrow 1$ just holds for $t \rightarrow 0$. In your solution $\dfrac{1}{x}$ doesn't satisfy the condition $\rightarrow 0$ but $\rightarrow \infty$.
A correct solution may be the following:
Since $\sin\dfrac{1}{x}$ is bounded and $x \rightarrow 0$, $\lim_{x \rightarrow 0}x \sin\frac{1}{x}=0$.
