I have the following function $f(x)$ defined as $x^4\sin(\frac{1}{x})$ if $x \neq 0$ and $0$ if $x=0$. And I'm asked if the function is:
a) differentiable
b) two times differentiable
c) two times continuously differentiable
This function is of course quite similar to the function $x^2\sin(\frac{1}{x})$, which can be found on the internet for the same type of exercises. But I still want to be sure that I proceeded in the correct way, because I have no solution to this exercise. Thanks for your feedback.
a) At $x=0$, we have $$\lim_{h\to 0}\frac{f(0+h)-f(0)}{h}=\lim_{h \to 0}\frac{h^4\sin(\frac{1}{h})}{h}=\lim_{h \to 0}h^3\sin(\frac{1}{h})=0$$ Everywhere else, we have simply $4x^3 \sin(\frac{1}{x})-x^2\cos(\frac{1}{x})$.
Thus the function is differentiable everywhere, even at $0$ where its value is $0$.
b) Again, if $x\neq 0$, we have simply $f^{\prime \prime}(x)=12x^2\sin(\frac{1}{x})-\sin(\frac{1}{x})-6x\cos(\frac{1}{x})$. At $x=0$, we have $$\lim_{h \to 0}\frac{f'(0+h)-f'(0)}{h}=\lim_{h \to 0}\frac{4h^3\sin(\frac{1}{h})-h^2\cos(\frac{1}{h})}{h}=\lim_{h \to 0}4h^2\sin(\frac{1}{h})-h\cos(\frac{1}{h})=0$$
So the function is also two times differentiable at $0$, where the derivative is $0$.
c) The function is however not two times continuously differentiable, because as said, the second derivative at $0$ is $0$, but if we take $\lim_{x \to 0_{\pm}} 12x^2\sin(\frac{1}{x})-\sin(\frac{1}{x})-6x\cos(\frac{1}{x})$, we don't always reach $0$ because $\sin(\frac{1}{x})$ oscillates between $-1$ and $1$. So the function is not two times continuously differentiable, even if the second derivative is defined everywhere, even at $0$
Are my results correct ?
Thanks for your help !
Your results are correct.
However your explanation on the fact that $12x^2\sin(\frac{1}{x})-\sin(\frac{1}{x})-6x\cos(\frac{1}{x})$ doesn't have a limit at $0$ is a bit fuzzy. For example $g(x)=\sin(\frac{1}{x})-\sin(\frac{1}{x})=0$ do have a limit at $0$ while $\sin(\frac{1}{x})$ don't. The sum of two maps that don't have a limit at a point may have a limit!
You should say that both $12x^2\sin(\frac{1}{x})$ and $6x\cos(\frac{1}{x})$ converges towards $0$ at $0$ while $\sin(\frac{1}{x})$ doesn't have a limit. Hence the sum of the three terms can't have a limit at zero.