Let
$$
f(x)=
\begin{cases}
x^2\cos\left(\frac{\pi}{x^2}\right) &\quad\text{if} \hspace{5mm} x\in (0,1]\\
0 &\quad\text{if} \hspace{5mm} \text{x = 0}
\end{cases} $$
Prove that the function is differentiable.
My try: I tried to use the limit to prove that the function is differentiable, but I didn't get anywhere, I think it's easier using "arguments", first of all I don't have any problems in $0$ beacuse the have the same value (it's continuous in $0$), and in the interval I want to say that decompostion is continuous because both are continuous, and the same argument for the product.
Do I need to use the fact that the function is differentiable at $a$ if $\lim_{h \rightarrow0}\frac{f(a+h)-f(a)}{h}$?
Or do you think it is enough to say that?
Best Answer
As we have $$\lim_{h \rightarrow 0^+}\frac{f(0+h)-f(0)}{h} =\lim_{h \rightarrow 0^+}h\cdot \cos\left(\frac{\pi}{h^2}\right)$$ then it's enough to find, that $\left| h\cdot \cos\left(\frac{\pi}{h^2}\right)\right| \leqslant |h|$.