Prove that, for any $x \in \mathbb{R}$ and $f \in (L^1\cap C)(\mathbb{R})$, $$\frac{1}{\epsilon}\int_{\mathbb{R}}f(t).\exp\left(\frac{-\pi(x-t)^2}{\epsilon^2}\right)dt \xrightarrow{\epsilon \to 0}f(x) $$
By substituting $u=\frac{\sqrt{\pi}}{\epsilon}(t-x)$, we get $du=\frac{\sqrt{\pi}}{\epsilon}dt$ and \begin{align*} \int_{\mathbb{R}}\exp\left(\frac{-\pi(x-t)^2}{\epsilon^2}\right)dt=\frac{\epsilon}{\sqrt{\pi}}\int_{\mathbb{R}}\exp\left(-u^2\right)du=\epsilon\\ \end{align*} Therefore \begin{align*} \frac{1}{\epsilon}\int_{\mathbb{R}}f(t).\exp\left(\frac{-\pi(x-t)^2}{\epsilon^2}\right)dt-f(x)\\ &=\frac{1}{\epsilon}\int_{\mathbb{R}}(f(t)-f(x))\exp\left(\frac{-\pi(x-t)^2}{\epsilon^2}\right)dt\\ &=\frac{1}{\sqrt{\pi}}\int_{\mathbb{R}}\left(f\left(\frac{\epsilon}{\sqrt{\pi}}u+x\right)-f(x)\right)\exp\left(-u^2\right)du\\ \end{align*} Suppose that $f$ is compactly supported. Let $[-a,a]$ be the compact support for $f$. Since $f$ is continuous, given $\eta > 0$, there exists $\delta > 0$ such that whenever $|h| < \delta$, we have $|f(x+h)-f(x)| < \eta$. For all $\epsilon < \frac{\sqrt{\pi}\delta}{2a}$, we have $\left|\frac{\epsilon}{\sqrt{\pi}}u+x-x\right|=\frac{\epsilon |u|}{\sqrt{\pi}} < \frac{\epsilon 2a}{\sqrt{\pi}} < \delta$ giving us $\left|f\left(\frac{\epsilon}{\sqrt{\pi}}u+x\right)-f(x)\right|< \eta$. Hence, for all $\epsilon < \frac{\sqrt{\pi}\delta}{2a}$ \begin{align*} \left|\frac{1}{\epsilon}\int_{\mathbb{R}}f(t).\exp\left(\frac{-\pi(x-t)^2}{\epsilon^2}\right)dt-f(x)\right|&\le \frac{1}{\sqrt{\pi}}\int_{\mathbb{R}}\left|\left(f\left(\frac{\epsilon}{\sqrt{\pi}}u+x\right)-f(x)\right)\right|\exp\left(-u^2\right)du\\ &\le\eta\frac{1}{\sqrt{\pi}}\int_{\mathbb{R}}\exp(-u^2)du=\eta\\ \end{align*} Now, since compactly supported functions are dense in $L^1$, for any function $f \in L^1$, there exists a compactly supported function $h $ such that $\|f-h\|_1 < \frac{\eta\sqrt{\pi}}{3}$. Therefore,
$$\left|\frac{1}{\epsilon}\int_{\mathbb{R}}f(t).\exp\left(\frac{-\pi(x-t)^2}{\epsilon^2}\right)dt-f(x)\right|$$ $$\le \frac{1}{\sqrt{\pi}}\int_{\mathbb{R}}\left|\left(f\left(\frac{\epsilon}{\sqrt{\pi}}u+x\right)-h\left(\frac{\epsilon}{\sqrt{\pi}}u+x\right)\right)\right|\exp\left(-u^2\right)du$$ $$+\frac{1}{\sqrt{\pi}}\int_{\mathbb{R}}\left|\left(h\left(\frac{\epsilon}{\sqrt{\pi}}u+x\right)-h(x)\right)\right|\exp(-u^2)du+\frac{1}{\sqrt{\pi}}\int_{\mathbb{R}}|h(x)-f(x)|\exp(-u^2)$$
I can make the first two integrals small. I am not able to make the third integral small. I am yet to use that $f$ is continuous. There is a way to do this problem using Dominated Convergence Theorem. I don't want to do that.
(Using Hint)
Let $g \in L^1(\mathbb{R})$. Let $E_n=[-n,n]$. Then $E_n \subset E_{n+1}$ for all $n$ and $\mathbb{R}=\cup_{n=1}^{\infty}E_n$. Therefore $$\int_{\mathbb{R}}|g|d\mu=\lim_ {n \to \infty}\int_{E_n}|f|d\mu$$ Let $\eta \gt 0$. Then there exists $n_0 \in \mathbb{N}$ such that for all $n \ge n_0$, we have $$\left|\int_{\mathbb{R}}|g|d\mu-\int_{E_n}|g|d\mu\right|\lt \eta \implies \left|\int_{|x| \gt n}|g(x)|d\mu \right| \lt \eta$$ Since $f, e^{-u^{2}} \in L^1(\mathbb{R})$, there exists $n_0 \in \mathbb{N}$ such that for all $n \ge n_0$, we have $\int_{|u| \gt n}|f(u)|du \lt \frac{\sqrt{\pi}\eta}{4}$ and $\int_{|u| \gt n}e^{-u^2}du \lt \frac{\sqrt{\pi}\eta}{4|f(x)|}$. `
Now $$\frac{1}{\sqrt{\pi}}\int_{\mathbb{R}}\left(f\left(\frac{\epsilon}{\sqrt{\pi}}u+x\right)-f(x)\right)\exp\left(-u^2\right)du=\frac{1}{\sqrt{\pi}}\int_{|u| \le \frac{1}{\sqrt{\epsilon}}}\left(f\left(\frac{\epsilon}{\sqrt{\pi}}u+x\right)-f(x)\right)\exp\left(-u^2\right)du+\frac{1}{\sqrt{\pi}}\int_{|u| \gt \frac{1}{\sqrt{\epsilon}}}\left(f\left(\frac{\epsilon}{\sqrt{\pi}}u+x\right)-f(x)\right)\exp\left(-u^2\right)du$$ Since $f$ is continuous at $x$, there exists a $\delta \gt 0$ such that whenever $|h| \lt \delta$, we have $|f(x+h)-f(x)| \lt \frac{\eta}{2}$. Thus, for $\epsilon \lt \pi\delta^2$, whenever $|u| \lt \frac{1}{\sqrt{\epsilon}}, |x+u\frac{\epsilon}{\sqrt{\pi}}-x| \lt \frac{\sqrt{\epsilon}}{\pi} \lt \delta$ and \begin{align*} \frac{1}{\sqrt{\pi}}\left|\int_{|u| \le \frac{1}{\sqrt{\epsilon}}}\left(f\left(\frac{\epsilon}{\sqrt{\pi}}u+x\right)-f(x)\right)\exp\left(-u^2\right)du\right|\\&\le \frac{1}{\sqrt{\pi}}\int_{|u| \le \frac{1}{\sqrt{\epsilon}}}\left|\left(f\left(\frac{\epsilon}{\sqrt{\pi}}u+x\right)-f(x)\right)\right|\exp\left(-u^2\right)du\\ &\le \frac{\eta}{2}\frac{1}{\sqrt{\pi}}\int_{|u| \le \frac{1}{\sqrt{\epsilon}}}e^{-u^2}du&(\text{ using continuity of f})\\ &\lt \frac{\eta}{2}\\ \end{align*} For $\epsilon \lt \frac{1}{n_0^2}, |u| \gt \frac{1}{\sqrt{\epsilon}} \gt n_0$ and $\left\{|u| \gt \frac{1}{\sqrt{\epsilon}}\right\} \subset \{|u| \gt n_0\}$. Therefore, \begin{align*} \frac{1}{\sqrt{\pi}}\left|\int_{|u| \gt \frac{1}{\sqrt{\epsilon}}}\left(f\left(\frac{\epsilon}{\sqrt{\pi}}u+x\right)-f(x)\right)\exp\left(-u^2\right)\right|du\\ &\le \frac{1}{\sqrt{\pi}}\int_{|u| \gt \frac{1}{\sqrt{\epsilon}}}\left|f\left(\frac{\epsilon}{\sqrt{\pi}}u+x\right)\right|du +\frac{1}{\sqrt{\pi}}|f(x)|\int_{|u| \gt \frac{1}{\sqrt{\epsilon}}}e^{-u^2}du\\ &\le\frac{1}{\sqrt{\pi}}\int_{|u| \gt n_0}\left|f\left(\frac{\epsilon}{\sqrt{\pi}}u+x\right)\right|du +\frac{1}{\sqrt{\pi}}|f(x)|\int_{|u| \gt n_0}e^{-u^2}du\\ &\le \frac{\eta }{2}\\ \end{align*} Thus, for $\epsilon \lt \min\{\pi\delta^2,\frac{1}{n_0^2}\}$, we have the claim.
Thanks for the help!!
Theorem: Let $g\in L^1_\text{loc}(\mathbb R)$ with $\int_{\mathbb R} g(x) dx = 1$. Then $\lim\limits_{\epsilon \to 0} \tfrac 1\epsilon g\big(\tfrac x\epsilon\big) = \delta(x)$
So in fact your integral is nothing but the convolution
$$\lim_{\epsilon \to 0} f* g_\epsilon = f * \delta = f$$
with $g(x) = e^{-\pi x^2}$. A proof of the theorem can be found here.