Let $u \in \mathcal{D}'(\mathbb{R})$. Show that:
$$\frac{u-\tau_{x}u}{x} \to \mathcal{D}u$$
in $\mathcal{D}'(\mathbb{R})$ when $x \to 0$. Here, we define $\tau_{s}f(x)=f(x+s)$.
It's an exercise from Rudin's Functional Analysis and I don't know how to start.
To follow the definitions of Rudin, I'll use the convention that $(\tau_x f)(y) = f(y-x)$. Writing $u'$ for the weak derivative of the distribution $u$, we need to show that for any test function $\varphi \in \mathcal{D}(\mathbb{R})$, $\langle \frac{u-\tau_x u}{x} , \varphi \rangle \to \langle u', \varphi \rangle$ as $x \to 0$. If $y$ denotes the variable in the duality pairing, then \begin{align*} \langle \frac{u-\tau_x u}{x} , \varphi \rangle &= \langle u(y) - (\tau_x u)(y), \frac{\varphi(y)}{x} \rangle\\ &= \langle u(y) , \frac{\varphi(y)}{x} \rangle - \langle u(y-x) , \frac{\varphi(y)}{x} \rangle\\ &= \langle u(y) , \frac{\varphi(y)}{x} \rangle - \langle u(y), \frac{\varphi(y+x)}{x} \rangle\\ &= -\langle u(y), \frac{\varphi(y+x) - \varphi(y)}{x} \rangle \to -\langle u(y), \varphi'(y) \rangle = \langle u', \varphi \rangle. \end{align*}