Deriving the Hessian from the limit definition of the derivative Could someone possibly help me understand how I can derive the Hessian matrix of a twice-differentiable function $f$ defined on $\mathbb{R}^n$ using the limit definition of the second derivative. Namely, how does: $\lim_{h -> 0}\frac{\nabla f(x+h) - \nabla f(x)}{h}$ result in the Hessian $\nabla^2 f(x)$. If I happen to be wrong about this, could you please point out what I am misunderstanding? 
Thank you very much! 

Let $f: \mathbb{R}^n \to \mathbb{R}$ be a differentiable function.
Then, at a point $p$, the derivative $Df\big|_p: \mathbb{R}^n \to \mathbb{R}$ can be computed by (but is not defined by)
$$
Df\big|_p(v) = \lim_{h \to 0} \frac{f(p+hv)-f(p)}{h}
$$
If $f$ is differentiable then $Df\big|_p$ is a linear function from $\mathbb{R}^n \to \mathbb{R}$.  We have that $f(p+v) \approx f(p)+Df\big|_p(v)$
If $f$ is twice differentiable, then we can think of its second derivative as a bilinear form $Hf\big|_p:\mathbb{R}^n \times \mathbb{R}^n \to \mathbb{R}$.  It can be computed by (but not defined by)
$$
Hf\big|_p(v,w) = \lim_{h \to 0} \frac{Df_{p+hv}(w) - Df\big|_p(w)}{h}
$$
We have that $Df\big|_{p+v}(w) \approx Df\big|_p(w)+Hf\big|_p(v,w)$.
It also turns out (the beginning of the multivariable Taylor's theorem), that $f(p+v) \approx f(p)+Df\big|_p(v)+\frac{1}{2!}Hf\big|_p(v,v)$.
The pattern continues with higher derivatives being higher order symmetric tensors.
