Show that $\int h_n'(x) \varphi(x)\, dx \to \langle \delta, \varphi\rangle$ - Generalized functions theory

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In the book Partial Differential Equations by Robert Strichartz, there's an exercise (#$1$, page $9$) that I am not quite sure how to solve. Is there anyone could give me the principal steps how to solve it?

Problem : Let $H$ be the distribution in $\mathbb{D}'(\mathbb{R})$ defined by the Heaviside function

$$ H(x) = \begin{cases} 1 & x>0\\ 0 & x \leq 0\\ \end{cases} $$

Show that if $h_n$ are differentiable functions such that $\int h_n \varphi(x) dx \to <H, \varphi>$ as $n \to \infty$ for all $\varphi \in \mathbb{D}$, then $$\int h_n'(x) \varphi(x)\, dx \to \langle \delta, \varphi\rangle.$$

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Using integration by parts we get:

$$\lim \limits _{n \to \infty} \int \limits _{-\infty} ^\infty h_n ' (x) \varphi (x) \ \textrm d x = \lim \limits _{n \to \infty} \left( h_n \varphi \Bigg|_{-\infty} ^\infty - \int \limits _{-\infty} ^\infty h_n (x) \varphi' (x) \ \textrm d x \right) = \\ \lim \limits _{n \to \infty} \left( (0 - 0) - \int \limits _{-\infty} ^\infty h_n (x) \varphi' (x) \ \textrm d x \right) = - \lim \limits _{n \to \infty} \langle h_n, \varphi' \rangle = - \langle H, \varphi' \rangle = - \int \limits _0 ^\infty \varphi' (x) \ \textrm d x = \\ -\big( \varphi (\infty) - \varphi (0) \big) = \varphi (0) = \langle \delta, \varphi \rangle ,$$

which shows that $\lim \limits _{n \to \infty} h_n ' = \delta$ (in the sense of distributions).

(Above, I have used twice that $\varphi (\infty) = \varphi (-\infty) = 0$, because $\varphi$ has compact support, which means that $\varphi$ is $0$ towards $\pm \infty$.)