# 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?

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Do you really need monotonicity to justify this? I mean, if $I(\epsilon)$ denotes the integral, then $\lim_{\epsilon\rightarrow 0^+}I(\epsilon)=I$ if and only if $I(\epsilon_n)$ tends to $I$ for every sequence $\epsilon_n$ tending to $0^+$. –  1015 Apr 28 '13 at 20:38