Uniform limit in definition of second order directional derivatives If $f:E\rightarrow F$ is twice differentiable at $x\in E$, do we then have $$\lim_{h,k\rightarrow 0}\frac{A_x(h,k)-f''(x)(h)(k)}{\|h\|\|k\|}=0$$ where $A_x(h,k):=f(x+h+k)-f(x+h)-f(x+k)+f(x)$?
This is a little stronger assumption than symmetry of second derivatives, also it implies that we can use a uniform limit in $$f''(x)(h)(k)=\lim_{\lambda\searrow 0}\lim_{\mu\searrow 0}\frac{A_x(\lambda h,\mu k)}{\lambda\mu}$$ (which is directly derived from the definition for directional derivatives) but I couldn't prove it yet.
Edit: H Cartan proves in Differential Calculus, Theorem 5.1.1, that $$\lim_{h,k\rightarrow 0}\frac{A_x(h,k)-f''(x)(h)(k)}{(\|h\|+\|k\|)^2}=0$$ and from this it follows e.g. that $$f''(x)(h)(k)=\lim_{\lambda\searrow 0}\frac{A_x(\lambda h,\lambda k)}{\lambda^2}$$ but I still couldn't derive the case of the uniform limit or the formula from the beginning.
 A: I set out to prove the statement $$\lim_{h,k\rightarrow 0}\frac{A_x(h,k)-f''(x)(h)(k)}{\|h\|\|k\|}=0.$$  Note that whenever this holds for $f$ at $x$ then it also holds for $f_{t}:=y\mapsto y+t$ at $x-t$ so we can w.l.o.g. assume $x=0$.  Also the statement is linear in the sense that if it holds for $f$, $g$ then also for $\lambda f+\mu g$ and it is not too difficult to verify that it holds for $x\mapsto f'(0)(x)+f''(0)(x)(x)$ (in fact we have equality for all $h,k$ in this case and don't even need to evaluate the limit), which is a mapping with the same first and second derivative at $0$ as $f$, so we can by substracting these terms safely assume $f'$ and $f''$ are both vanishing at $0$.
But this means that for any $\epsilon>0$ there is $\delta>0$ such that $\|f'(x)\|\le\|x\| \epsilon$ whenever $\|x\|\le\delta$.  Now fix $\epsilon$ and a corresponding $\delta$ and let $\|h\|+\|k\|\le\delta$, we can w.l.o.g assume $\|h\|\le\|k\|$.  Define $$B_x(h):=\frac{f(x+h)-f(x)}{\|h\|},$$ by the mean value theorem we know $\|B_x(h)\|\le\sup_{0\le\lambda\le 1}\|f'(x+\lambda h)\|$ so $$\|B_k(h)-B_0(h)\|\le\|B_k(h)\|+\|B_0(h)\|\le(2\|h\|+\|k\|)\epsilon$$ and further $$\frac{\|A_0(h,k)\|}{\|h\|\|k\|}=\frac{\|B_k(h)-B_0(h)\|}{\|k\|}\le\frac{(2\|h\|+\|k\|)\epsilon}{\|k\|}\le3\epsilon.$$
