Q:

Prove that $\lim_{k \to \infty} \int_{-\infty}^{\infty} \frac{ke^{-k^2x^2}}{\pi}h(x) = h(0)$

Prove that  
$$\lim_{k \to \infty} \int_{-\infty}^{\infty} \frac{ke^{-k^2x^2}}{\pi}h(x) \, dx= h(0) \tag{A}$$ 
Proof 
From the left side of (A)  
$$ \frac{k}{\pi}\lim_{k \to \infty} \int_{-\infty}^{\infty} e^{-k^2x^2}h(x)\, dx$$    
Integration by parts? 
$$ \frac{k}{\pi}\lim_{k \to \infty} \int_{-\infty}^{\infty} e^{-k^2x^2}h(x)\, dx =  
\frac{k}{\pi}\lim_{k \to \infty} \left[
\left( \int e^{-k^2x^2} \right)h(x)\bigg|_{-\infty}^{\infty} +
 \int_{-\infty}^{\infty} \left( \int e^{-k^2x^2} \right)h'(x)\, dx \right]$$   
But here there's no antiderivative for the Gaussian-type function, unless I can apply those limits of integration to it, but I don't see how.  The definite integral of that piece would be $\sqrt{\pi}/{k}.  $   
Then we'd have  
$$ \frac{k}{\pi}\lim_{k \to \infty} \int_{-\infty}^{\infty} e^{-k^2x^2}h(x)\, dx =  
\frac{k}{\pi}\lim_{k \to \infty} \left[
\frac{\sqrt{\pi}}{{k}}h(x)\bigg|_{-\infty}^{\infty} +
 \int_{-\infty}^{\infty} \left( \int e^{-k^2x^2} \right)h'(x)\, dx \right]$$  
Don't know where to go from here.  The actual question is to prove that the kernel in the integrand in (A) is Dirac Delta sequence.  

A:

If $h\in C^{0}(-\infty,\infty)\cap L^{\infty}(-\infty,\infty)$, then 
\begin{align*}
\dfrac{1}{\sqrt{\pi}}\int_{-\infty}^{\infty}e^{-u^{2}}(h(u/k)-h(0))du\rightarrow 0,~~~~k\rightarrow\infty
\end{align*}
because $h(u/k)\rightarrow h(0)$ for every $u$ as $k\rightarrow\infty$ and $|h(u/k)-h(0)|\leq 2\|u\|_{L^{\infty}(-\infty,\infty)}$ and $e^{-u^{2}}\in L^{1}(-\infty,\infty)$, so Lebesgue Dominated Convergence Theorem. 
Now use the fact that $\displaystyle\int_{-\infty}^{\infty}e^{-u^{2}}du=\sqrt{\pi}$, we have 
\begin{align*}
\dfrac{1}{\sqrt{\pi}}\int_{-\infty}^{\infty}e^{-u^{2}}h(u/k)du\rightarrow h(0),~~~~k\rightarrow\infty. 
\end{align*}
For the question, let $u=kx$ for the substitution.  If the denominator in question is $\pi$, then it may not go through.