Define $g(x) = \begin{cases}|x|^\alpha\cos(1/x^2) & x\neq 0\\ 0 & x = 0\end{cases}.$
Determine what values of $\alpha>0$ is $g(x)$ differentiable at $x = 0$.
I had this question on an assignment and I got it wrong. Apparently the answer is $\alpha > 1$, however, I am having trouble understanding this, can someone help me?
In approaching this question, I determined that first I need continuity at $x = 0$. Therefore, I need a value for $\alpha$, such that $\lim_{x \to 0} |x|^\alpha\cos(1/x^2) = 0.$
So for this requirement, I need $\alpha\geq 1$. Is that correct? Because if I just test $(0.001)^{0.001}$, this gives something like $0.99$.
I will also need to ensure that $g'(x)$ from the left and right to be equal, so that there are no corners, or cusps at $x = 0$ (i.e. $g'(0) = 0$, since $\cos$ and $\sin$ oscillate).
Since $\cos(1/x^2)$ and $\sin(1/x^2)$ oscillate between $-1$ and $1$, I will need $2|x|^{\alpha - 3} \to 0$, as $x \to 0$, in order to have both the left and right derivatives the same.
So for the $(\alpha - 3)$, I would need this to be $\geq 1$. Therefore $\alpha > 4$.
Can someone help me to understand what I am getting wrong here?