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

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.

• $\frac{nx}{nx^2}=\frac{1}{x}$, not $\frac{1}{nx}$. – carmichael561 May 13 '17 at 3:32
• Also $g(x)$ cannot be a dominating function since it is not integrable on $(0, 1]$. – Sangchul Lee May 13 '17 at 3:37

$$\int_{[0,1]} \frac{nx}{1+nx^2} \, dx = \int_{[0,1]} \frac{x}{\frac{1}{n}+x^2} \, dx.$$
It is easy to check that the integrand is monotone increasing in $n$. So by the monotone convergence theorem,
$$\lim_{n\to\infty} \int_{[0,1]} \frac{nx}{1+nx^2} \, dx = \int_{[0,1]} \lim_{n\to\infty} \frac{x}{\frac{1}{n}+x^2} \, dx = \int_{[0,1]} \frac{1}{x} \, dx = \infty.$$
• This works;so (+1). But isn't it trivial to see that $\int_0^1 \frac{nx}{1+nx^2}\,dx=\frac12\log(1+n)\to \infty$? – Mark Viola May 13 '17 at 3:55