I'm not sure how to approach this problem:
Find the value of : $\lim_{n \rightarrow \infty} \int_{0}^{1+ \frac{1}{n}} n \cdot \log(1+ \frac{x^2}{n}) dx$.
I think I'm supposed to solve this using Beppo-Levi or dominated Convergence theorem.
If I could show that $\left|\mathbb{1}_{(0,1+ \frac{1}{n})}(x) \cdot n \cdot \log\left(1+ \frac{x^2}{n}\right)\right| ≤ x^2$, one condition for dCt would be met.
1) How can I show that the left side is always smaller than its limit? I this even true for all $x \in (0,1+ \frac{1}{n})$? Would this even suffice, because I can't show the inequality for the limit $\mathbb{1}_{(0,1)}(x) \cdot x^2$?
2) How do I show that $n \cdot \log\left(1+ \frac{x^2}{n}\right)$ is integrable, in the sense that the integral over its absolute value is finite?
You can use that $t-t^2\le \log(1+t)\le t$ for all $t\ge 0$.
Then $x^2-\frac{x^4}{n}\le n\log(1+\frac{x^2}{n})\le x^2$. You can use dominated convergence to show the convergence of the integrals if you like, but as the sequence of functions converges uniformly, you can do without.
To see that $t-t^2\le \log(1+t)\le t$, note that $1+t\ge t$ and $(1-2t)(1+t)=1-t-t^2\le 1$ so $1-2t\le \frac1{1+t}\le 1$. Integrating this inequality gives you the wanted inequality, as $0-0^2=\log(1+0)=0$.