This question is related to the square of Dirac delta distribution. In one of the answers to this question, it was argued that if we represent $\delta(x)$ as a limit of a rectangular peak with unit area, width $\epsilon$, and height $1/\epsilon$, we obtain $$ \lim_{\epsilon \to 0} \int_{0}^{\epsilon} [\delta_{\epsilon}(x)]^{2} \mathrm{d}{x} = \lim_{\epsilon \to 0} \frac{1}{\epsilon}.$$
So, my question is, given a smooth function $f$ satisfying the following properties (so that $\lim_{\epsilon \to 0} f'(x) = \delta(x)$) $$f(x) = \begin{cases} 1 &\text{for } x \geq \epsilon \\ h(x) &\text{for } 0 < x < \epsilon \\ 0 &\text{for } x \leq 0 \end{cases}$$ where $h(x)$ is some function interpolating the two asymptotic values, can we show that $$ \lim_{\epsilon \to 0} \int_{0}^{\epsilon} [f'(x)]^{2} \mathrm{d}{x} = \lim_{\epsilon \to 0} \frac{1}{\epsilon}$$ is true without specifying the form of $h(x)$ explicitly?
I think I've figured out how to show that this is the case (assuming the family of smooth functions considered is constructed via uniform dilation).
Consider a family of smooth functions $$ f_{\epsilon}(x) = \begin{cases} 1 &\text{for } x \geq \epsilon \\ g_{\epsilon}(x) &\text{for } 0 < x < \epsilon \\ 0 &\text{for } x \leq 0 \end{cases} \implies f'_{\epsilon}(x) = \begin{cases} 0 &\text{for } x \geq \epsilon \\ g'_{\epsilon}(x) &\text{for } 0 < x < \epsilon \\ 0 &\text{for } x \leq 0 \end{cases} $$ where $g_{\epsilon}(x)$ is some smooth interpolating function and prime indicates derivative with respect to the argument. When we fix $ \epsilon = r $, we have $$ f_{r}(x) = \begin{cases} 1 &\text{for } x \geq r \\ g_{r}(x) &\text{for } 0 < x < r \\ 0 &\text{for } x \leq 0 \end{cases} \implies f'_{r}(x) = \begin{cases} 0 &\text{for } x \geq r \\ g'_{r}(x) &\text{for } 0 < x < r \\ 0 &\text{for } x \leq 0 \end{cases} $$ and $$ \int_{-\infty}^{+\infty} \left[ f'_{r}(x) \right]^{p} \mathrm{d}x = \int_{0}^{r} \left[ g'_{r}(x) \right]^{p} \mathrm{d}x \equiv K_{r}^{p} < \infty$$ for any non-negative integer $p$ since $ g'_{r}(x) $ is necessarily bounded. Otherwise, it is not smooth. Note that some care is required if one is to extend $p$ to rational or real numbers. Note also that, by definition, there is no dependence on $\epsilon$ in $K_{r}^{p}$ as we have already fixed $\epsilon = r$.
Now, we know that $f_{\epsilon}(x)$ is just a strectched version of $f_{r}(x)$. So, we can write $$ f_{\epsilon}(x) = f_{r}(sx) $$ for some stretching function $s$ which, in general, depends on $x$ and $\epsilon$. In our case, however, we would like the stretching to be uniform for all $x$. Thus, we require that $s$ to be independent of $x$. Given the criteria, we have, in particular, that $$ f_{\epsilon}(\epsilon) = f_{r}(s\epsilon) = f_{r}(r). $$ It follows that $s = r/\epsilon$ and $$ f'_{\epsilon}(x) = \frac{r}{\epsilon} f'_{r} \left( \frac{rx}{\epsilon} \right). $$
Thus, \begin{align} \int_{0}^{\epsilon} [g'_{\epsilon}(x)]^{p} \mathrm{d}x = \int_{-\infty}^{+\infty} [f'_{\epsilon}(x)]^{p} \mathrm{d}x &= \left( \frac{r}{\epsilon} \right)^{p} \int_{-\infty}^{+\infty} \left[ f'_{r}\left(\frac{rx}{\epsilon}\right) \right]^{p} \mathrm{d}x \\ &= \left( \frac{r}{\epsilon} \right)^{p} \int_{0}^{\epsilon} \left[ g'_{r}\left(\frac{rx}{\epsilon}\right) \right]^{p} \mathrm{d}x \\ &= \left( \frac{r}{\epsilon} \right)^{p-1} \int_{0}^{r} [g'_{r}(y)]^{p} \mathrm{d}y \\ &= \left( \frac{r}{\epsilon} \right)^{p-1} K_{r}^{p} \end{align} or, written compactly \begin{align} K_{\epsilon}^{p} &= \left( \frac{r}{\epsilon} \right)^{p-1} K_{r}^{p} \\ \implies \lim_{\epsilon \to 0} K_{\epsilon}^{p} &= \lim_{\epsilon \to 0} \left( \frac{r}{\epsilon} \right)^{p-1} K_{r}^{p}. \end{align} In particular, we have the desired result (except for some pre-factor independent of $\epsilon$) when $ p = 2$, that is $$ \lim_{\epsilon \to 0} \int_{0}^{\epsilon} [f'_{\epsilon}(x)]^{2} \mathrm{d}x = \lim_{\epsilon \to 0} \frac{r}{\epsilon} K_{r}^{2}. $$