Calculate $\lim_{n\to\infty}\int_{[0,1]}\frac{nx}{1+nx^2}$.

Question

It is required to calculate $\lim_{n\to\infty}\int_{[0,1]}\frac{nx}{1+nx^2}$. Here is my attempt.

Let $x\in(0,1]$. Then $|\frac{nx}{1+nx^2}|\leq|\frac{nx}{nx^2}|=\frac{1}{nx}$ for each $n\in\mathbb{N}$. Hence it follows that $f_n(x)=\frac{nx}{1+nx^2}$ converges pointwise to $f\equiv 0$. Moreover $|f_n(x)|\leq g(x)$, where $g(x)=\frac{1}{x}$ if $x\in(0,1]$ and $g(0)=0$. So $|f_n(x)|\leq g(x)$ $a.e.$ on $[0,1]$ and $g$ is integrable as it is continuous $a.e.$ on $[0,1]$. Now by Lebesgue Dominated Convergence theorem $\lim_{n\to\infty}\int_{[0,1]}f_n=\int_{[0,1]}f\ $ , i.e. $\lim_{n\to\infty}\int_{[0,1]}\frac{nx}{1+nx^2}=0$.

Could someone please tell me if this solution is alright? Thanks.

Answer 1

As pointed out by comments, your attempt has a couple of issues. Here is a valid approach. Write

$$ \int_{[0,1]} \frac{nx}{1+nx^2} \, dx = \int_{[0,1]} \frac{x}{\frac{1}{n}+x^2} \, dx. $$

It is easy to check that the integrand is monotone increasing in $n$. So by the monotone convergence theorem,

$$ \lim_{n\to\infty} \int_{[0,1]} \frac{nx}{1+nx^2} \, dx = \int_{[0,1]} \lim_{n\to\infty} \frac{x}{\frac{1}{n}+x^2} \, dx = \int_{[0,1]} \frac{1}{x} \, dx = \infty. $$