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Consider $\lim_{n \to \infty} \int_0^1 f_n$ where $f_n$ takes $$ \text{(a)} \frac{nx \log(x)}{1+n^2x^2} \text{and}$$ $$ \text{(b)} \frac{n^{3/2}x}{1+n^2x^2} \text{and}$$

Attempt: For the part (a) $$\frac{nx \log(x)}{1+n^2x^2} \leq \frac{nx \log(x)}{2nx} = \frac{\log(x)}{2} \in L^1$$

Using the above dominating function I evaluated the limit as $0$. I am not able to figure out a dominating function for the part (b). I also wanted to know if what is done for part (a) is correct.

Thanks in advance.

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  • $\begingroup$ your inequality is wrong the sign $\leq$ must be $\geq$ $\endgroup$
    – Akmal
    Feb 26, 2013 at 4:07
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    $\begingroup$ I think it is right $1+n^2 x^2 \geq 2nx$. You reciprocate them. $\endgroup$
    – user62089
    Feb 26, 2013 at 4:08
  • $\begingroup$ Yes, you are right, but if $x\in (0, 1]$ then $nx\log x$ is negative isn't? $\endgroup$
    – Akmal
    Feb 26, 2013 at 9:31
  • $\begingroup$ @DonAntonio: Dear, I absolutely do not agree with you. If you do not believe this, check one more time "when we can use inequalities" $\endgroup$
    – user64144
    Feb 27, 2013 at 2:37

2 Answers 2

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For the second one you don't even need the DCT. Directly:

$$\int\limits_0^1\frac{n^{3/2}x}{1+n^2x^2}dx=\frac{1}{2\sqrt n}\int\limits_0^1\frac{2n^2x\,dx}{1+n^2x^2}=\left.\frac{1}{2\sqrt n}\log(1+n^2x^2)\right|_0^1=$$

$$=\frac{1}{2\sqrt n}\log(1+n^2)\xrightarrow[n\to\infty]{}0$$

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Inspired by the first example, I think AM-GM inequality also works for the second one.

$(x_1+x_2+…+x_n)/n\geq(x_1x_2…x_n)^{1/n}$

To produce $n^{3/2}$ on RHS, let's consider inequality with 4 $x$s, which is implied by $\frac 3 2=\frac 6 4$.

Then we have

$1+n^{2}x^3=1+3(\frac{1}{3}n^{2}x^3)\geq4(\frac{1}{27}n^6x^9)^{1/4}$

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