How to solve $\lim_{t \to 0}\frac{\frac{1}{|a+th|}(a+th) - \frac{1}{|a|}a}{t}$ Title really says it all.
I'm trying to compute this limit. It is the differential of the function $f(x) =  \frac{x}{|x|}$ in the direction $h$.
So in our limit $\lim_{t \to 0}\frac{\frac{1}{|a+th|}(a+th) - \frac{1}{|a|}a}{t}$ we have that $a, h \in \mathbb R^n, |a| \neq 0, |h| = 1$, $t \in \mathbb R$.
This should be equal to $\frac{h}{|a|} - \frac{a \langle a, h\rangle}{|a|^3}$ if the answer provided is to be believed.
We could try to verify this by actually computing $\frac{df_i}{dt}(a+th)$ at $t = 0$ but that hardly seems pleasant.
 A: If you define the function $\;f:\Bbb R^n\to\Bbb R^n\,,\,\,f(x)=\frac x{\left\|x\right\|}\;$ , we then have
$$f(x)=\left(\frac{x_1}{\sqrt{\sum\limits_{k=1}^n x_k^2}}\,,\,\,\frac{x_2}{\sqrt{\sum\limits_{k=1}^n x_k^2}}\,,\ldots,\,\frac{x_n}{\sqrt{\sum\limits_{k=1}^n x_k^2}}\right)\implies$$$${}$$
$$\nabla f_x=\left(\frac{\sum\limits_{k\neq 1}^nx_k^2}{\left(\sum\limits_{k=1}^n x_k^2\right)^{3/2}}\,,\,\,\frac{\sum\limits_{k\neq 2}^nx_k^2}{\left(\sum\limits_{k=1}^n x_k^2\right)^{3/2}},\,\ldots,\,\frac{\sum\limits_{k\neq n}^nx_k^2}{\left(\sum\limits_{k=1}^n x_k^2\right)^{3/2}}\right)$$
Observe all the partial derivatives of first order exist and are continuous in some neighborhood of $\;a\;$ , thus the directional derivative at $\;a\;$ in the direction of $\;h\;$ is given by derivative is given by
$$\nabla f_a\cdot\frac h{\left\|h\right\|}$$
Try to take it from here.
A: Note that by binomial series
$$|a+th|=\langle a+th, a+th\rangle^\frac12=(|a|^2+2t\langle a, h\rangle+t^2)^\frac12=\\=|a|\left(1+\frac{2t\langle a, h\rangle}{|a|^2}+\frac{t^2}{|a|^2}\right)^\frac12=|a|+\frac{t\langle a, h\rangle}{|a|}+o(t^2)$$
thus
$$\frac{\frac{1}{|a+th|}(a+th) - \frac{1}{|a|}a}{t}=\frac{|a|(a+th)-|a+th|a}{t|a||a+th|}=\frac{|a|a+t|a|h-|a|a-\frac{t\langle a, h\rangle}{|a|}a+o(t^2)}{t|a|^2+o(t^2)}=\frac{|a|h-\frac{\langle a, h\rangle}{|a|}a+o(t)}{|a|^2+o(t)}\to \frac{h}{|a|}-\frac{\langle a, h\rangle}{|a|^3}a$$
