When does $ |{x}|^\alpha\sin{\frac1{x}} $ have continuous derivative at zero?

Question

Let

$$f(x) :=\begin{cases} |{x}|^\alpha\sin{\frac1{x}},&x\ne0\\0,&\text{otherwsie}\end{cases}$$

I've come up with that the derivative at zero is:

$$\lim_{x\to0}\frac{|{x}|^\alpha\sin{\frac1{x}}}{x} = 0$$ for $\alpha\ge2$.

But how do I get the derivative function to see when it's equal to its limit?

$$\lim_{x\to y}\frac{|{x}|^\alpha\sin{\frac1{x}}-|{y}|^\alpha\sin{\frac1{y}}}{x-y} = \text{the function}$$

Or how do I compute $$\lim_{y\to0}\lim_{x\to y}\frac{|{x}|^\alpha\sin{\frac1{x}}-|{y}|^\alpha\sin{\frac1{y}}}{x-y} = \text{desired limit to compare with }0$$

Answer 1

Your function doesnt have a simple way to define it derivative directly from the definition of derivative at a point because it is a composition of functions. In this case it is better to use the chain and the product rule, that show that

$$[(f\circ g)(x)\cdot h(x)]'=[(f\circ g)(x)]'\cdot h(x)+(f\circ g)(x)\cdot h'(x)\\=(f'\circ g)(x)g'(x)h(x)+(f\circ g)(x)h'(x)$$

whenever the functions involved are differentiable at the point $x$. In your case, for $x\neq 0$, we have that

$$f'(x)=[|x|^\alpha\sin(1/x)]'=[|x|^\alpha]'\sin(1/x)+|x|^\alpha[\sin(1/x)]'\\=\alpha|x|^{\alpha-1}\operatorname{sign}(x)\sin(1/x)+|x|^\alpha\cos(1/x)\cdot\frac{-1}{x^2}\\=|x|^{\alpha-2}(\alpha x\sin(1/x)-\cos(1/x))$$

where $\operatorname{sign}$ is the sign function. From here you only need to see for what values of $\alpha$ it holds that

$$\lim_{x\to0}f'(x)=0$$

Answer 2

use that $$\frac{|x|^{\alpha}|\sin(\frac{1}{x})|}{|x|}\le |x|^{\alpha-1}$$ it must be $$\alpha>1$$ so $$|x|^{\alpha-1}$$ tends to zero for $x$ tends to zero