Let $$D^+u(x) = \left\{v \in \mathbb{R}; \limsup_{y\to x}\frac{u(y)-u(x)-v\cdot (y-x)}{|y-x|} \le0\right\}$$ and
$$D^-u(x) = \left\{v \in \mathbb{R}; \liminf_{y\to x}\frac{u(y)-u(x)-v\cdot (y-x)}{|y-x|} \ge0\right\}$$ the superdifferential and the subdifferential of the function $u :O \to \mathbb{R}$ continuous with $O \subset \mathbb{R}$, where $ u(x) = \begin{cases} 0 & \text{if } x < 0 \\ \sqrt{x} & \text{if } x\in [0,1]\\ 1 & \text{if } x > 1 \end{cases}$
I have to show that :
$$D^+u(0)=\varnothing$$ $$D^-u(0)=[0,\infty)$$
$$D^+u(x) = D^-u(x) = \left\{\frac{1}{2\sqrt(x)}\right\}$$
$$D^+u(1) = [0,1/2]$$ $$D^-u(1)=\varnothing$$
I don't know anything about the superdifferential and subdifferentiel except that it's the set of supergradient and subgradient. Is there a general method to show these 5 questions ?