Take the normal definition of step function $\theta(x)$ and delta function $\delta(x)$ such that $$ \frac{d}{dx}\theta(x)=\delta(x). $$
I conjecture that $$ \lim_{n \to 0^+} \frac{n \theta(x)}{x^{1-n}}=\delta(x). $$
Can we prove or disprove this?
Then how about $$ \lim_{n \to 0^-} \frac{n \theta(x)}{x^{1-n}}=? $$
As $\eta\to 0^+$ (dont call it $n$ please)
$$x^{\eta}1_{x >0}\to 1_{x >0}$$ in $L^1_{loc}$ and thus the sense of distributions, therefore its distributional derivative
$$\eta x^{\eta-1} 1_{x >0} = (x^{\eta}1_{x >0})' \to (1_{x >0})' = \delta$$
in the sense of distributions.
$$\eta x^{-1-\eta} 1_{x >0}$$ is not $L^1_{loc}$, it is not a distribution.
We still have $$x^{-\eta}1_{x >0}\to 1_{x >0}, \qquad (x^{-\eta}1_{x >0})' \to (1_{x >0})' = \delta$$ but $(x^{-\eta}1_{x >0})'$ is not $-\eta x^{-1-\eta} 1_{x >0}$, it is $-\eta x^{-1-\eta} 1_{x >0}$ minus some kind of regularization term around $0$ making it a distribution.