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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.

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  • $\begingroup$ $\frac{nx}{nx^2}=\frac{1}{x}$, not $\frac{1}{nx}$. $\endgroup$
    – carmichael561
    Commented May 13, 2017 at 3:32
  • $\begingroup$ Also $g(x)$ cannot be a dominating function since it is not integrable on $(0, 1]$. $\endgroup$
    – Sangchul Lee
    Commented May 13, 2017 at 3:37

1 Answer 1

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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,

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

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    $\begingroup$ This works;so (+1). But isn't it trivial to see that $\int_0^1 \frac{nx}{1+nx^2}\,dx=\frac12\log(1+n)\to \infty$? $\endgroup$
    – Mark Viola
    Commented May 13, 2017 at 3:55
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    $\begingroup$ @Dr.MV Oh, yes. Your right, exact computation is always a nice way if available. I acknowledge that I did not think about that option when I first saw this. On the other hand, this answer might be also relevant to OP as it seems an analysis exercise. $\endgroup$
    – Sangchul Lee
    Commented May 13, 2017 at 3:59
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    $\begingroup$ I suspect that you're correct given the tags. $\endgroup$
    – Mark Viola
    Commented May 13, 2017 at 4:00

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