Evaluation of a limit with integral Is this limit
$$\lim_{\varepsilon\to 0}\,\,\varepsilon\int_{\mathbb{R}^3}\frac{e^{-\varepsilon|x|}}{|x|^2(1+|x|^2)^s}$$
with $s>\frac{1}{2}$, zero?.
The limit of a product is the product of limit, so I evaluate
$$\lim_{\varepsilon\to 0}\,\,\int_{\mathbb{R}^3}\frac{e^{-\varepsilon|x|}}{|x|^2(1+|x|^2)^s}$$.
With the theorem of dominated convergence the last limit equals
$$\int_{\mathbb{R}^3}\frac{1}{|x|^2(1+|x|^2)^s}=4\pi\int_{0}^{+\infty}\frac{1}{(1+r^2)^s}=C<\infty$$
(I have used the fact that $s>\frac{1}{2}$)
Using the product rule I have the result.
Have I made some mistake?

What you did is correct. The only thing you have to take care is that in general, dominated convergence theorem applies for sequences. Here there is no problem since the convergence is monotonic. 
