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
 A: $$\lim_{x\to0}\frac{x^2\sin\frac1x}{\sin x}=\lim_{x\to0}\frac{x\sin\frac1x}{\frac{\sin x}x}=\lim_{x\to0}\frac{x\sin\frac1x}1=0.$$
A: 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 $x \rightarrow 0$, $\lim_{x \rightarrow 0}x \sin\frac{1}{x}=0$.
A: we need to show that for any positive $\epsilon$ we can provide a positive $\delta$ so that:
$$ \left\lvert \frac{x^2 \, \sin \frac{1}{x}}{\sin x} \right\rvert < \epsilon \,\,\,\,\,\, \Longleftarrow \,\,\,\,\,\,0 < |x| < \delta.$$
assume that $\,\,\delta < \pi$, $\,\,$ so that we can be sure $\sin x \neq 0$. Then, for any $x$ satisfying $0 < |x| < \delta$,
$$ \left\lvert \frac{x^2 \, \sin \frac{1}{x}}{\sin x} \right\rvert \,\,\,=\,\,\, \left\lvert \frac{x^2}{\sin x} \right\rvert \left\lvert \sin \frac{1}{x}   \right\rvert  \,\,\,\leq\,\,\, \left\lvert \frac{x^2}{\sin x} \right\rvert .$$
Therefore, it is good enough to show that
$$  \lim_{x\to 0} \, \frac{x^2}{\sin x} \,\,=\,\,0. $$
But to show that is an easy L'hopital's rule exercise. Then choose $\delta$ the smaller of $\pi$ and the $\delta_1$ from that easy limit.
Remark - this should work for $\tan x$ or $\sinh x$ in place of $\sin x$. In the case of tangent, require $\,\,\delta < \frac{\pi}{2}.$
