Q:

If $f$ is differentiable at the point $\hat{x}$, show that $f$ possesses directional derivative in all directions $v \in \mathbb{R}^n - \{0\}$.

Let $U \subset \mathbb{R}^n$ an open set, $\hat{x} \in U$ and $f : U
 \to \mathbb{R}$. We define the directional derivative of $f$ in the
  direction of $v \in \mathbb{R}^n - \{0\}$ at the point $\hat{x}$ by
  the following limit if it exists : $$D_v f(\hat{x})= \lim_{h \to 0}
 \frac{f(\hat{x}+hv)-f(\hat{x})}{h}.$$ If $f$ is differentiable at the
  point $\hat{x}$, show that $f$ possesses directional dirivative in all
  directions $v \in \mathbb{R}^n - \{0\}$.

Then we have there exist $f'(\hat{x}) \in \mathcal{L}(\mathbb{R}^n,\mathbb{R})$ such that $$\lim_{x \to \hat{x}} \frac{f(x)-f(\hat{x})-f'(\hat{x})(x-\hat{x})}{\|x-\hat{x} \|}=0.$$
Let $x=\hat{x}+hv$.
$$\lim_{\hat{x}+hv \to \hat{x}} \frac{f(\hat{x}+hv)-f(\hat{x})-f'(\hat{x})(\hat{x}+hv-\hat{x})}{\|\hat{x}+hv-\hat{x} \|}=\lim_{hv \to 0} \frac{f(\hat{x}+hv)-f(\hat{x})-f'(\hat{x})(hv)}{\|hv \|}= 0$$
I know I can almost show, but I do not know how to fill in the rest. Does someone could guide me through the rest of the resolution of this question?

A:

If $v = 0$, it is trivial. Then suppose $v \neq 0$. If $f$ is differentiable at $\hat{x}$, then we can write $$f(\hat{x}+hv) - f(\hat{x}) = df(\hat{x})(hv)+r(hv),$$where $r(hv)/\|hv\|\to 0$ as $h \to 0$. Then: \begin{align}D_vf(\hat{x}) &= \lim_{h \to 0}\frac{f(\hat{x}+hv)-f(\hat{x})}{h} \\ &= \lim_{h \to 0} \frac{df(\hat{x})(hv)+r(hv)}{h} \\ &= df(\hat{x})(v) + \require{cancel}\cancelto{0}{\lim_{h \to 0} \|v\| \frac{r(hv)}{h\|v\|}} \\ &= df(\hat{x})(v). \end{align}

