$h_n$ is a function such that for every test $\varphi$ is $\int_{-\infty}^{+\infty}h_n(x) \varphi (x)dx\rightarrow H(\varphi).$ Then $\int_{-\infty}^{+\infty}h'_n(x) \varphi (x)dx\rightarrow \varphi(0)$.
I would start with the following facts
(1) $H(x)=\int_{-\infty}^x \delta(t)dt$ then $\frac{dH}{dx}=\frac{d}{dx}\int_{-\infty}^x \delta(t)dt=\delta(x)$
(2) The general Heaviside distribution is given by $\int_{-\infty}^{+\infty}H(x) \varphi (x)dx=\int_{0}^{+\infty} \varphi (x)dx$
Now replace $f$ by $h_n$, since $h_n'$ exists $\frac{d}{dx}\int_{-\infty}^{+\infty}h(x) \varphi (x)dx\rightarrow \frac{d}{dx} H(x)=\delta(x)$
I know that if $\delta[x]=\int_{-\infty}^{+\infty}f(x)\varphi(x)dx$ then $\delta[\varphi]=\varphi(0)$, because f this it seems that the former limit shouldn't be $\delta(x)$ but $\delta (\varphi)$.
Where is it wrong?
Using integration by parts, properties of the test functions, the assumption of the problem, and FTC, we have $$ \lim_{n\to \infty}\int_{-\infty}^\infty h'_n(x)\varphi(x) dx = \lim_{n\to \infty}-\int_{-\infty}^\infty h_n(x)\varphi'(x)dx = -\int_0^\infty\varphi'(x)dx = \varphi(0)$$