I am trying to prove that for a smooth function $f\in C^\infty$ which decays suitably fast, we have $$\frac{1}{\epsilon}\int_{-\infty}^\infty\frac{f(x)}{\frac{1}{\epsilon^2}+(x-\frac{1}{\epsilon})^2}\,dx=\pi f(\frac{1}{\epsilon})+O(\epsilon).$$
Start of proof:
We use that $\int_{-\infty}^\infty\frac{a}{a^2+x^2}\,dx=\pi$ for $a>0$.
This gives us
\begin{align}\frac{1}{\epsilon}\int_{-\infty}^\infty\frac{f(x)}{\frac{1}{\epsilon^2}+(x-\frac{1}{\epsilon})^2}\,dx&=\frac{1}{\epsilon}\int_{-\infty}^\infty\frac{f(\frac{1}{\epsilon})}{\frac{1}{\epsilon^2}+(x-\frac{1}{\epsilon})^2}\,dx+\frac{1}{\epsilon}\int_{-\infty}^\infty\frac{(f(x)-f(\frac{1}{\epsilon}))}{\frac{1}{\epsilon^2}+(x-\frac{1}{\epsilon})^2}\,dx\\ &=\pi f(\frac{1}{\epsilon})+\frac{1}{\epsilon}\int_{-\infty}^\infty\frac{(f(x)-f(\frac{1}{\epsilon}))}{\frac{1}{\epsilon^2}+(x-\frac{1}{\epsilon})^2}\,dx.\end{align}
So the rest of the proof is to show that $$\frac{1}{\epsilon}\int_{-\infty}^\infty\frac{(f(x)-f(\frac{1}{\epsilon}))}{\frac{1}{\epsilon^2}+(x-\frac{1}{\epsilon})^2}\,dx$$ is $O(\epsilon)$ as $\epsilon\rightarrow 0^+$.
Equivalently, following the hint given by @J.C., the statement is equivalent to proving that $$\int\frac{f(x)-f(\frac{1}{\epsilon})}{1+(1-x\epsilon)^2}\,dx=O(1).$$
To see this, we divide the real line into a set where $x\sim\frac{1}{\epsilon}$ and a set where $x\not\sim\frac{1}{\epsilon}$.
On the set $|x-\frac{1}{\epsilon}|\leq\frac{1}{2\epsilon}$ the fact that $f\in C^\infty$ and the assumption that $f$ decays sufficiently fast allows us to Taylor expand around $\frac{1}{\epsilon}$. We assumee $$f(x)=f(\frac{1}{\epsilon})+(x-\frac{1}{\epsilon})O(\epsilon)^2$$ for $x\sim \frac{1}{\epsilon}$ is valid. Then $$\Bigg|\int_{\{|x-\frac{1}{\epsilon}|\leq\frac{1}{2\epsilon}\}}\frac{f(x)-f(\frac{1}{\epsilon})}{1+(1-x\epsilon)^2}\,dx\Bigg|=\Bigg|\int_{\{|x-\frac{1}{\epsilon}|\leq\frac{1}{2\epsilon}\}}\frac{(x-\frac{1}{\epsilon})O(\epsilon)^2}{1+(1-x\epsilon)^2}\,dx\Bigg|\leq\frac{1}{\epsilon}\frac{\frac{1}{2\epsilon}O(\epsilon^2)}{1}=O(1)$$ using the crude estimate $|\int_I f(x)\,dx|\leq |I|\sup_{x\in I}|f(x)|$.
On the set $|x-\frac{1}{\epsilon}|>\frac{1}{2\epsilon}$ we use that $f$ is uniformly bounded: \begin{align}\Bigg|\int_{\{|x-\frac{1}{\epsilon}|>\frac{1}{2\epsilon}\}}\frac{f(x)-f(\frac{1}{\epsilon})}{1+(1-x\epsilon)^2}\,dx\Bigg|&\leq 2\|f\|_{L^\infty}\int_{\{|x-\frac{1}{\epsilon}|>\frac{1}{2\epsilon}\}}\frac{1}{1+(1-x\epsilon)^2}\,dx\\ &=2\|f\|_{L^\infty}\int_{\{|y|>\frac{1}{2}\}}\frac{1}{1+y^2}\,\frac{dy}{\epsilon}\end{align} where we made the change of variables $y=\epsilon x-1$. This means I can only show that $$\int\frac{f(x)-f(\frac{1}{\epsilon})}{1+(1-x\epsilon)^2}\,dx=O(\frac{1}{\epsilon}).$$