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Find $$\lim_{n \to \infty} \int_{1}^{n}\frac{nx^{1/2}}{1+nx^2}dx$$

I have tried tackling this problem using the DCT but I am not quite sure if I have the right answer. To begin with, $$\int_{1}^{n}\frac{nx^{1/2}}{1+nx^2}dx=\int_{1}^{\infty}\frac{nx^{1/2}}{1+nx^2}1_{[1,n]}dx$$

I think that , $$\bigg|\frac{n\sqrt{x}}{1+nx^2}\bigg|\leq x^{-3/2}$$

Now applying the DCT gives $$\lim_{n \to \infty} \int_{1}^{n}\frac{nx^{1/2}}{1+nx^2}dx=2$$ but I am not sure if the procedure that I followed is correct. I would appreciate any help.

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    Hint: Use a substitution $x = t/\sqrt{n}$ first. – 2017-02-12

2 Answers 2

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Yes, the pointwise limit of the $f_n$'s will be $f(x) = \frac{1}{x^{3/2}}$, and this will also serve as an (integrable) dominating function on $(1, \infty)$. Thus the dominated convergence theorem implies $$\lim_{n \to \infty}\int_1^\infty \frac{nx^{1/2}}{1 + nx^2}\chi_{[1, n]}(x) dx = \int_1^\infty \lim_{n \to \infty}\frac{nx^{1/2}}{1 + nx^2}\chi_{[1, n]}(x) dx = \int_1^\infty \frac{1}{x^{3/2}}dx = 2.$$

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    I answered more or less at the same time, but your answer says a bit more than mine. (+1) – 2017-02-12
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    Yes, I noticed! (And frankly I agree with what you wrote - there wasn't much that needed saying.) – 2017-02-12
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    Thank you very much for your very informative answer. – 2017-02-12
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$$\lim_{n\to +\infty}\int_{1}^{n}\frac{n\sqrt{x}}{1+nx^2}\,dx = \lim_{n\to +\infty}\left(\int_{1}^{+\infty}\frac{\sqrt{x}}{\frac{1}{n}+x^2}\,dx-\int_{n}^{+\infty}\frac{\sqrt{x}}{\frac{1}{n}+x^2}\right)$$ where $$ \left|\int_{n}^{+\infty}\frac{\sqrt{x}}{\frac{1}{n}+x^2}\right|\leq \int_{n}^{+\infty}x^{-3/2}\,dx=\frac{2}{\sqrt{n}} $$ hence the limit is given by $$ \lim_{n\to +\infty}\int_{1}^{+\infty}\frac{\sqrt{x}}{\frac{1}{n}+x^2}\,dx \stackrel{\text{DCT}}{=}\int_{1}^{+\infty}x^{-3/2}\,dx = \color{red}{2}.$$