Question: 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.

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 $g(x) = \dfrac{1}{2\sqrt{x}}$.
For $|f\_{n}(x)|=|\dfrac{nx\log(x)}{1+n^{2}x^{2}}|\leq |\dfrac{\log(x)}{2}|\leq |\log(x)|$. So for $g(x) = \log(x)$. Is that right??