Vtyjkyuk

We define the function $f$ in the following manner:

$$f(x) =
begincases
0, & textif $x = 0$ \
vert x vert^alpha textsin(frac1x), & textif $x neq 0$
endcases$$

Prove that the function $f$ is differentiable at $x=0$ if and only if $alpha > 1$

($Rightarrow$) Suppose the function $f$ is differentiable at $x=0$, then the limit:
$$lim_x rightarrow 0 fracf(x)-f(0)x-0=lim_x rightarrow 0 fracvert xvert^alpha textsin(frac1x)x$$

must exist, I claim that the limit exists and is equal to $0$ when $alpha > 1$

Case (1): $x>0$, we have:
$$lim_x rightarrow 0 x^alpha-1 textsin(frac1x)$$

Idea: Prove that $lim_x rightarrow 0 x^alpha-1 =0$ if and only if $alpha > 1$, then apply Squeeze Theorem.

Hence for every $epsilon>0$, choose $delta=epsilon ^ frac1alpha - 1$(This can be done because $alpha > 1$), then:
$$0<vert xvert<delta Rightarrowvert xvert^alpha -1<delta^alpha - 1=epsilon$$

Hence by Squeeze Theorem, I must have the limit being equal to $0$.

Case (2) is similar except that it is for $x<0$

Not sure if I missed out anything though, appreciate it if you can provide me with some feedback on this proof!

Any idea on how to do $(Leftarrow)$ is greatly appreciated, or is it actually trivial? (As in I can deduce it from ($Rightarrow$))

To show $leftarrow$ note that $x^alpha-1$ diverges as $xto 0^+$ and $alpha<1$. Also $sindfrac1x$ has no limit when $xto 0$ therefore the limit $x^alpha -1sindfrac1x$ doesn't exist if $alpha le 1$. So it should be that $alpha >1$. If so we can write $$lim_xto 0^+x^alpha-1sin dfrac1x=lim_xto inftydfracsin xx^alpha-1=0$$this means that your approach is correct. The same is true for $x<0$ since the function is odd.