= Solution
The <L1-normalized dilation of a distribution> is
$$
\boxed{\langle u_\varepsilon,\phi\rangle=\langle u,\phi(\varepsilon\,\cdot)\rangle.}
$$
For a <regular distribution> this is the <function> $u_\varepsilon(x)=\varepsilon^{-1}u(x/\varepsilon)$. In terms of the wiki's <dilation of a distribution> convention, $u_\varepsilon=\varepsilon^{-1}\delta_{1/\varepsilon}u$; the normalization preserves the <L1 space> norm.
If $u\in L^1(\mathbb R)$, the <dominated convergence theorem>, with dominating <integrable function> $|u(y)|\|\phi\|_\infty$, gives
$$
\langle u_\varepsilon,\phi\rangle=\int u(y)\phi(\varepsilon y)\,dy
\longrightarrow\phi(0)\int u(y)\,dy.
$$
Consequently
$$
\boxed{u_\varepsilon\longrightarrow\left(\int_{\mathbb R}u\right)\delta_0
\quad\text{in }\mathcal D'(\mathbb R).}
$$
An arbitrary <distribution> need not have this behavior. Take the constant <regular distribution> $u=1$. Then $u_\varepsilon=\varepsilon^{-1}$, so for a <test function> with integral one,
$$
\langle u_\varepsilon,\phi\rangle=\varepsilon^{-1}\longrightarrow+\infty.
$$
Hence $\boxed{u=1\text{ gives no distributional limit}}$.
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