I have tried to use the Lebesgue Dominated Convergence Theorem to evaluate:
$$\lim_{n\rightarrow \infty} \int_{(0,1]} f_n \;d\mu $$ with $f_n(x)=\dfrac{n\sqrt{x}}{1+n^2x^2}$ and $f_n(x)=\dfrac{n\;x\log(x)}{1+n^2x^2}$.
The problem is I can't find the function $g$ such that $\lvert f_n(x) \rvert \leq g(x)$ with $g\in L^1$. $g(x)=\dfrac{\lvert \log(x) \rvert}{x}$ and $g(x)=\dfrac{1}{x^{3/2}}$ is the best I got and these functions are not Lebesgue integrable.
Best Answer
For $|f_n(x)| = \left|\dfrac{nx \log (x)}{1+n^2x^2}\right| \leq 2|\log x| = -2\log x, x \in (0,1]$. Take $g(x) = - 2 \log x$.
For $|f_n(x)| = \left|\dfrac{n\sqrt{x}}{1+n^2x^2}\right| = \left|\dfrac{nx}{1+n^2x^2}\right|\cdot \dfrac{1}{\sqrt{x}} \leq \dfrac{1}{2\sqrt{x}}$. Take $g(x) = \dfrac{1}{2\sqrt{x}}$.