For $b\in \mathbb{R}$, let $f$ be defined by $ f(x) = \left\{ \begin{array}{c} x^b \sin\frac{1}{x}, &x > 0 \\ 0, &x \le 0 \end{array} \right. $ For $b\in \mathbb{R}$, let $f$ be defined by
$$
f(x) = 
\left\{
\begin{array}{c}
x^b \sin\frac{1}{x}, &x > 0 \\ 
0, &x \le 0
\end{array}
\right.
$$
Prove the following:


*

*$f$ is continuous at $0 \iff b > 0$

*$f$ is differentiable at $0 \iff b > 1$

*$f'$ is continuous at $0 \iff b > 2$


I'm not really sure how to prove the if and only if parts. For example in part 1, would I just prove it with each case being $b > 0$, $b = 0$, then $b < 0$?
 A: Hint: For "if", relate the questions to limits of the expressions below and remember the squeeze theorem and that $|\sin x|\leq 1$, $|\cos x|\leq 1$ for all $x$:


*

*$x^b\sin \left(1/x\right) $

*$\frac{x^b\sin \left(1/x\right)}{x} = x^{b-1}\sin \left(1/x\right)$

*$bx^{b-1}\sin \left(1/x\right) - x^{b-2}\cos \left(1/x\right)$
For "only if", exploiting the same expressions, use sequences $\frac{1}{2n\pi +\pi/2}$ and $\frac{1}{2\pi n}$ (converging to $0$) to show lack of continuity/differentiability at $0$ of $f$  (and $f'$) when $b$ does not respect the wanted conditions.
A: Hints.


*

*If $b>0$, then $\lvert f(x)\rvert\le \lvert x\rvert^b$, and hence $\lim_{x\to 0}f(x)=0=f(0)$.

*If $b>1$, then $\dfrac{f(h)-f(0)}{h}=\dfrac{h^b\sin (1/h)}{h}=h^{b-1}\sin(1/h)\to 0$, as $h\to 0$, since  $\lvert h^{b-1}\sin(1/h)\rvert\le \lvert h\rvert^{b-1}\to 0$.

*If $b>2$, then $f'(x)=bx^{b-1}\sin(1/x)-x^{b-2}\cos(1/x)$, for $x\ne 0$, while $f'(0)=0$. And
$$
\vert f'(x)\rvert\le b\lvert x\rvert^{b-1}+\lvert x\rvert^{b-2}\to 0=f'(0).
$$
