For $\alpha \in \mathbb{R}$ define $f:\mathbb{R} \longrightarrow \mathbb{R}$ as:
$ \begin{cases} 0 & \text{if } x=0 \\ |x|^\alpha sin(\frac{1}{x})& \text{if } x \ne0 \\ \end{cases} $
For the first three parts, state (by proving), the values of $\alpha$ for which:
$(a)$ $f$ is continuous at $0$
$(b)$ $f$ is differentiable at $0$
$(c)$ $f$ is continuously differentiable at $0$ (differentiable with the derivative continuous at $0$)
Clearly show detailed working and mention any theorems you use.
A solution I found is as follows:
$(a)$ $lim_{x\rightarrow 0} f(x)=f(0) =0$. Therefore, $f$ is continuous $\forall \alpha >0$
$(b)$ Let $\phi(x)=\frac{f(x)-f(0)}{x-0}=\frac{f(x)}{x}=\frac{x^{\alpha}sin(\frac{1}{x})}{x}$
$\phi(x)=|x|^{\alpha-1}sin(\frac{1}{x})$
Hence, $f$ is differentiable if $\alpha -1>0; \alpha>1$
$(c)$ $f$ is differentiable and its derivative is continuous if $\alpha>2$
It claims to be a complete solution but I am not sure whether it is complete and rigorous. Any help with explanation and verification would be appreciated.
We know that $\lim_{x\to 0}|x|^\alpha$ exists iff $\alpha>0$ and then limit is $0$.
$\left||x|^\alpha sin(\frac{1}{x})\right|\le|x|^\alpha $, for all $ x\ne0$. $\implies \lim_{x\to 0}\left||x|^\alpha sin(\frac{1}{x})\right|\le\lim_{x\to 0}|x|^\alpha=0$ for $\alpha>0$
$\left||x|^\alpha cos(\frac{1}{x})\right|\le|x|^\alpha $, for all $ x\ne0$. $\implies \lim_{x\to 0}\left||x|^\alpha cos(\frac{1}{x})\right|\le\lim_{x\to 0}|x|^\alpha=0$ for $\alpha>0$
So for $\alpha>0$ $$ \lim_{x\to 0}f(x)=\lim_{x\to 0}|x|^\alpha sin(\frac{1}{x})=0 $$Hence f is continuous at $0$. If you take $\alpha<0$ take the sequence {$f(\frac{2}{n\pi})$} this unbounded hence $\lim_{x\to 0}f(x)$ doesn't exist.
This answers (a).
In the similar way also try to prove the same thing for $g(x)=\begin{cases} 0 & \text{if } x=0 \\\\ |x|^\alpha cos(\frac{1}{x})& \text{if } x \ne0 \\ \end{cases}$
For (b) is direct application of (a).
$\phi(x)=\frac{f(x)-f(0)}{x-0}=\frac{f(x)}{x}=\frac{x^{\alpha}sin(\frac{1}{x})}{x}=|x|^{\alpha-1}sin(\frac{1}{x}),\forall x>0$
$\phi(x)=\frac{f(x)-f(0)}{x-0}=\frac{f(x)}{x}=\frac{|x|^{\alpha}sin(\frac{1}{x})}{x}=-|x|^{\alpha-1}sin(\frac{1}{x}),\forall x<0$
For (c) I need to show $$\lim_{x\to 0}f'(x)=f'(0)=0$$ By chain rule you have $$f'(x)=\alpha x^{\alpha-1}\cdot \sin(\frac{1}{x})-x^{\alpha}\left(\frac{-1}{x^2}\right)\cos\left(\frac{1}{x}\right)=\alpha x^{\alpha-1}sin(\frac{1}{x})+x^{\alpha-2}cos(\frac{1}{x}) \text{, for } x>0$$ so $\lim_{x\to 0+} f'(x)$ exists iff $\alpha-2>0$. and if this limit exists it is equal to $0$.
Now observe $f$ is an odd function (ie $f(-x)=-f(x)$). Then you have $f'(-x)=-f'(x)\implies\lim_{x\to 0-} f'(x)=\lim_{x\to 0+}f'(-x)=\lim_{x\to 0+}-(f'(x))=0$
So you have $$\lim_{x\to 0}f'(x)=f'(0)=0$$.