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Calculate

$\lim_{n \rightarrow \infty}$ $\int_1^\infty$ $\sqrt{x} \over {1 + nx^3}$ $d\lambda(x)$.

My attempt:

First, we note that $\sqrt{x} \over {1 + nx^3}$ is integrable as a composition of the integrable functions $\sqrt{x}$ and ${1 + nx^3}.$ It is obvious that

$\lim_{n \rightarrow \infty}$ $\sqrt{x} \over {1 + nx^3}$ $= 0 =: f$.

Furthermore, since ${1 + nx^3} \ge \sqrt{x} \ \ \forall n \ge 1, \ x\in [1, \infty)$, we know that

$\sqrt{x} \over {1 + nx^3}$ $\le 1 =: M$, with $M$ being (trivially) an integrable function on $[1, \infty)$.

Hence, we can apply the dominated convergence theorem, which yields

$\lim_{n \rightarrow \infty}$ $\int_1^\infty$ $\sqrt{x} \over {1 + nx^3}$ $d\lambda(x) = \int_1^\infty 0 d\lambda = 0.$

2 Answers 2

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Unless $\lambda$ is a finite measure on $(1,\infty)$, the function $1$ is not integrable. You can always improve your bound by saying $1+nx^3 >x^3$.

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    In order to conclude $\sqrt{x} \over 1+nx^3$ $<$ $\sqrt{x} \over x^3$? – 2017-01-21
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    yes because even with, say the standard Lesbesgue measure, $\frac{\sqrt{x}}{x^3}=x^{-2.5}$ is still integrable. – 2017-01-21
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    So that would be enough to conclude the rest as I did above? – 2017-01-21
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    Yes, you still want to use the dominated convergence theorem to arrive at your conclusion that the limit is 0. – 2017-01-21
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Since $M:=1$ isn't integrable on $[1,\infty)$, your approach doesn't work. Your result, however, is correct, as I show below.

By substituting $t:=\sqrt[3]{n}x$, we have $dx=\frac{dt}{\sqrt[3]{n}}$ and $\sqrt{x}=\frac{\sqrt{t}}{\sqrt[6]{n}}$. Also, $t(1)=\sqrt[3]{n}$. Since $\sqrt[3]{n}\cdot\sqrt[6]{n}=\sqrt{n}$, we have

\begin{align} \lim\limits_{n\rightarrow\infty}\int_1^\infty\frac{\sqrt{x}}{1+nx^3}dx&=\lim\limits_{n\rightarrow\infty}\frac{1}{\sqrt{n}}\int_{\sqrt[3]{n}}^\infty\frac{\sqrt{t}}{1+t^3}dt \end{align}

Now, as $\sqrt[3]{n}>1$ for $n>1$ and $\int_1^\infty\frac{\sqrt{t}}{1+t^3}dt$ exists (this is easily shown), we know that the integral remains finite for any value of $n$.

Thus, in the limit, the product tends to $0$ because $\frac{1}{\sqrt{n}}\rightarrow 0$.