Can we show $\lim_{\varepsilon\to0}\sup_{0<\left\|x-y\right\|_E<\varepsilon}\frac{|f(x)-f(y)|}{\left\|x-y\right\|_E}=\left\|{\rm D}f(x)\right\|_{E'}$? Let $f:E\to\mathbb R$ be Fréchet differentiable and $x\in E$. By definition, $$\lim_{y\to x}\frac{|f(x)-f(y)-{\rm D}f(x)(x-y)|}{\left\|x-y\right\|_E}=0.\tag1$$ Can we somehow (maybe by imposing a further suitable assumption) show that $$\lim_{\varepsilon\to0}\sup_{\substack{y\in E\\0<\left\|x-y\right\|_E<\varepsilon}}\frac{|f(x)-f(y)|}{\left\|x-y\right\|_E}=\left\|{\rm D}f(x)\right\|_{E'}\tag2?$$
 A: If $E$ is a Banach space, and $u \colon E \to \mathbb{R}$ is a
bounded linear functional, then for all $\delta > 0,$
$$
\|u\| = \sup_{\|x\| = 1}|u(x)| =
\sup_{\|h\| = \delta}
\left\lvert{u}\left(\frac{h}{\|h\|}\right)\right\rvert =
\sup_{\|h\| = \delta}\frac{|u(h)|}{\|h\|}.
$$
Therefore for all $\varepsilon > 0,$
\begin{equation}
\label{3708359:eq:3}\tag{3}
\|u\| = \sup_{0 < \|h\| < \varepsilon}\frac{|u(h)|}{\|h\|}.
\end{equation}
If $U$ is an open subset of $E,$ and $f \colon U \to \mathbb{R}$ is
differentiable at $x \in U,$ then the function
$$
d \colon U - x \to \mathbb{R}, \
h \mapsto f(x + h) - f(x) - f'(x)(h)
$$
satisfies the condition
$$
\lim_{h \to 0}\frac{d(h)}{\|h\|} = 0.
$$
Take any $\eta > 0,$ and take $\varepsilon > 0$ such that if
$\|h\| < \varepsilon$ then $x + h \in U$ and
$$
|d(h)| \leqslant \eta\|h\|.
$$
If $x + h \in U,$ then by the Triangle Inequality,
$$
||f(x + h) - f(x)| - |f'(x)(h)|| \leqslant
|f(x + h) - f(x) - f'(x)(h)| = |d(h)|.
$$
Therefore, if $0 < \|h\| < \varepsilon,$
$$
\left\lvert \frac{|f(x + h) - f(x)|}{\|h\|} -
\frac{|f'(x)(h)|}{\|h\|} \right\rvert \leqslant \eta.
$$
Take any $\varepsilon^*$ such that $0 < \varepsilon^* < \varepsilon,$
and let
$$
N(\varepsilon^*) = \{h \in E : 0 < \|h\| < \varepsilon^* \}.
$$
For $h \in N(\varepsilon^*),$ the inequality just proved has the form
$$
|p(h) - q(h)| \leqslant \eta,
$$
for certain functions
$$
p, q \colon N(\varepsilon^*) \to \mathbb{R}_{\geqslant 0}.
$$
By \eqref{3708359:eq:3}, the function $q(h) = |f'(x)(h)|/\|h\|$ is bounded on $N(\varepsilon^*),$ and
$$
\sup_{h \in N(\varepsilon^*)}q(h) = \|f'(x)\|.
$$
Therefore, the function $p(h) = |f(x + h) - f(x)|/\|h\|$ is also bounded
on $N(\varepsilon^*),$ and
$$
\left\lvert \sup_{0 < \|h\| < \varepsilon^*}
\frac{|f(x + h) - f(x)|}{\|h\|} - \|f'(x)\| \right\rvert
\leqslant \eta.
$$
This holds for arbitrary $\eta$ and for all $\varepsilon^*$ such that
$0 < \varepsilon^* < \varepsilon,$ where $\varepsilon$ depends on $\eta,$
therefore
$$
\lim_{\varepsilon \to 0+} \sup_{0 < \|h\| < \varepsilon}
\frac{|f(x + h) - f(x)|}{\|h\|} = \|f'(x)\|.
$$
A: Let $\{v_n\}$ be such that $\lim_{k\rightarrow \infty} \|Df(x)(v_n)\|=\|Df(x)\|$ and $\|v_n\|=1$ for all $n$. Then, by definition, choosing $t_n \rightarrow 0$, we have
$$\lim_{n \rightarrow \infty}\frac{|f(x+t_n v_n)-f(x)|}{|t_n|}=\lim_{n\rightarrow \infty}|Df(x)(v_n)|=\|Df(x)\|.$$ Hence, 
$$\lim_{\varepsilon\to0}\sup_{\substack{y\in E\\0<\left\|x-y\right\|_E<\varepsilon}}\frac{|f(x)-f(y)|}{\left\|x-y\right\|_E} \geq \left\|{\rm D}f(x)\right\|.$$ Suppose, by contradiction, that 
$$\lim_{\varepsilon\to0}\sup_{\substack{y\in E\\0<\left\|x-y\right\|_E<\varepsilon}}\frac{|f(x)-f(y)|}{\left\|x-y\right\|_E} > \left\|{\rm D}f(x)\right\|.$$ Then, choosing $\{u_n\}\subset E$ and $\{t_n\}$ such that $\|u_n\|=1$ for all $n$, $t_n\rightarrow 0$ and
$$\lim_{\varepsilon\to0}\sup_{\substack{y\in E\\0<\left\|x-y\right\|_E<\varepsilon}}\frac{|f(x)-f(y)|}{\left\|x-y\right\|_E} = \lim_{k\rightarrow \infty} \frac{|f(x+t_n u_n)-f(x)|}{|t_n|},$$ we get
$$\|Df(x)(u_n)\| = \dfrac{\|\left( f(x+t_n u_n)- f(x)\right)-\left(-Df(x)(t_n u_n)+ f(x+t_n u_n)-f(x)\right)\|}{|t_n|} \geq  $$
$$ \dfrac{\|f(x+t_n u_n)-f(x)\| - \| f(x+t_n u_n)- f(x)-Df(x)(t_n u_n)\|}{|t_n|}.$$ Finelly, for large enough $n$,
$$
\|Df(x)(u_n)\|> \|Df(x)\|.
$$ A contradiction.
