How to check if $f(x)=x^2\sin(x)^{-1}$ is differentiable at 0?

Question

How would I check if $f(x)=x^2\sin(x)^{-1}$ is differentiable at 0?

I tried using the definition of a derivative but got stuck at the limit part. Does one care what happens at the limit point, or only what happens near it?

\begin{align} f^\prime(0) &= \lim_{x\to0}\dfrac{f(x)-f(0)}{x-0}\\ &= \lim_{x\to0}\dfrac{x^2\sin(x)^{-1}-0}{x-0}\\ &= \lim_{x\to0}\dfrac{x}{\sin(x)}\\ &=1? \text{ or does it not exist?} \end{align}

Answer 1

Note that

$$\lim_{x\to 0} \frac{x^2}{\sin x}=\lim_{x\to 0} x\cdot \frac{x}{\sin x} =0 $$

and since $f(x)$ is not defined at $x=0$ we need to define $f(0)=0$ to make it countinuous at that point. Then, as you have shown, since by the definition the limit exists, $f(x)$ is differentiable at $x=0$.

Answer 2

Doing a simple Taylor expansion works well:

When $x$ approches $0$, $\frac{x}{\sin(x)} = \frac{x}{x - \frac{x^3}{6}+\text{o}(x^3)} = \frac{1}{1 - \frac{x^2}{6}+\text{o}(x^2)} \rightarrow 1$.

So, yeah

$ \begin{align} f^\prime(0) &= \lim_{x\to0}\dfrac{x}{\sin(x)} =1. \end{align} $