Let $ f:\mathbb{R}^m \rightarrow \mathbb{R}^n$ differentiable at $x$ and $v$ an unit vector in $\mathbb{R}^m$. I've always used the formula that the directional derivative in $x$ by $v$ is simply the Jacobian times the $v$. However, I could never figure it out how to prove it.
$$\frac{\partial f}{\partial v}(x) = Df(x) v$$
In this question, the answer simply assumes that
$$\lim_{t\to0}\frac{\bigl\lVert f(a+tv)-f(a)-Df(a)(tv)\bigr\rVert}{\lVert tv\rVert}=\lim_{t\to0}\frac{f(a+tv)-f(a)-tDf(a)(v)}t$$
But it apparently ignores the triangular inequality in the given norm.
What am I not seeing here? How can I prove it?
It just uses that $\Vert tv\Vert=\vert t\vert\Vert v\Vert$, so we have
$$\frac{\Vert f(a+tv)-f(a)-\mathrm Df(a)(tv)\Vert}{\Vert tv\Vert}=\frac{1}{\Vert v\Vert}\left\Vert\frac{f(a+tv)-f(a)-\mathrm Df(a)(tv)}{t}\right\Vert.$$
Now the condition is that the limit of this expression is $0$. This means that
$$\lim_{t\to0}\left\Vert\frac{f(a+tv)-f(a)-\mathrm Df(a)(tv)}{t}\right\Vert=0,$$
since $\frac{1}{\Vert v\Vert}$ is constant. And a function goes to zero if and only if its norm goes to zero, since $f(x)\to a$ means $\Vert f(x)-a\Vert\to0$ by definition. So this is equivalent to
$$\lim_{t\to0}\frac{f(a+tv)-f(a)-\mathrm Df(a)(tv)}{t}=0.$$