Find $\lambda$ such that $f$ it is differentiable in zero and has a continuous derivative in zero

Question

I am trying to solve this task

Find $\lambda>0$ such that $f=\begin{cases}0& x=0\\ |x|^{\lambda}\cdot \sin\frac{1}{x} & x\neq 0 \end{cases}$

a) is differentiable in zero

b) has a continuous derivative in zero.

My try

a) $$\lim_{h \rightarrow 0^+}\frac{f(h)-f(0)}{h} = \lim_{h \rightarrow 0^+}\frac{|h|^{\lambda}}{h} \cdot \sin\frac{1}{h} = 0$$

$$\lim_{h \rightarrow 0^-}\frac{f(h)-f(0)}{h} = \lim_{h \rightarrow 0^+}\frac{|h|^{\lambda}}{h} \cdot \sin\frac{1}{h} = 0$$ But there I have some doubts. It seems like $\lambda$ does not matter there....
Moreover, can somebody give me hints to b)?
I think that I have do the same thing but this is unbelievable

Answer 1

For the existence part, to make sure the limit $$ f'(0)=\lim_{h\to 0}\ \frac{|h|^{\lambda}}{h}\sin\frac{1}{h} $$ exists, we have $ \lambda>1 $. And the limit is $ 0 $.

For the continuous derivative part, we have to let the following limit equal to $ 0 $, i.e. : $$ \lim_{h\to 0}\ f'(h)-f'(0)=\lim_{h\to 0}\ f'(h)-0=\lim_{h\to 0}\ f'(h)=0 .$$ Thus, $$ \begin{align} \lim_{h\to 0+}f'(h)&= \lim_{h\to 0+}\left[(\lambda-1)h^{\lambda-2}\sin\frac{1}{h}-h^{\lambda-3}\cos\frac{1}{h}\right]=0 \end{align} $$ and $$ \begin{align} \lim_{h\to 0-}f'(h)&= -\lim_{h\to 0-}\left[(-\lambda+1)(-h)^{\lambda-2}\sin\frac{1}{h}-(-h)^{\lambda-3}\cos\frac{1}{h}\right]=0 \end{align} .$$ Therefore, $ \lambda> 3 $.