We know the following: if $f \colon X \to Y$ is directionally differentiable and Lipschitz continuous then it is Hadamard differentiable. We also know that the Hadamard derivative is continuous wrt the direction (see this answer).
What I don't understand is the function $f(x) = \max(0,x)$ satisfies these conditions yet the derivative at $0$ is not continuous. We have $$f'(0)(d) = \begin{cases} 1 &: d > 0\\ 0 &: d <= 0 \end{cases} $$ which is not continuous in $d$ (take $d_n = (-1)^n/n$ for example). What am I missing?
Here I use the following definitions:
$f$ is directionally differentiable at $x$ in direction $h$ if there is a function $f'(x)\colon X\rightarrow Y$ such that \begin{align} \lim_{t\rightarrow0+}\frac{f(x+th)-f(x)}{t}-f'(x)(h)=0 \end{align}
$f$ is Hadamard differentiable at $x$ in direction $h$ if there is a function $f'(x):X\rightarrow Y$ such that for any $h_t\xrightarrow{t\rightarrow0+}h$
\begin{align}
\lim_{t\rightarrow0+}\frac{f(x+th_t)-f(x)}{t}-f'(x)(h)=0
\end{align}