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In preparation for finals, I am trying to calculate $$\lim_{n\rightarrow \infty} \int\limits_0^\infty \frac{n^2 \sin(x/n)}{n^3x + x(1 + x^3)} \, d x$$ with proof.

Here is my approach/what I have done so far: If we can find a dominating function, we have $$\lim_{n\rightarrow \infty} \int\limits_0^\infty \frac{n^2 \sin(x/n)}{n^3x + x(1 + x^3)} \, d x = \int\limits_0^\infty \lim_{n\rightarrow \infty} \frac{n^2 \sin(x/n)}{n^3x + x(1 + x^3)} \, d x$$ by the Dominated Convergence Theorem. If we let $f_{n} = \frac{n^2 \sin(x/n)}{n^3x + x(1 + x^3)}$, then $f_{n}(x)$ converges to $0$ for all $x > 0$, which implies the limit is equal to 0 because the Dominated Convergence Theorem only requires a.e. convergence (so not having convergence at $x = 0$ is no issue). Operating under the assumption that dominating function exists, is this correct?

As far as finding a dominating function is concerned, we have $$ |f_{n}| = \left| \frac{n^2 \sin(x/n)}{n^3x + x(1 + x^3)} \right| = \frac{|n^2 \sin(n/x)|}{n^3x + x(1 + x^3)} \leq \frac{n^2}{n^3x + x(1 + x^3)}, $$ which is where I get stuck. The two directions that seemed the most clear from here was to either $$ \frac{n^2}{n^3x + x(1 + x^3)} \leq\frac{n^2}{n^3x} = \frac{1}{x} \quad \text{or} \quad \frac{n^2}{n^3x + x(1 + x^3)} \leq \frac{n^2}{x(1 + x^3)}. $$ The former is not integrable and I cannot seem to grapple with the $n^2$ on the latter and sufficiently bound it. So my main question is how can I bound $|f_{n}|$?

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  • $\begingroup$ Mathematica yields $0$. $\endgroup$ – David G. Stork Dec 6 '17 at 1:43
  • $\begingroup$ Desmos yields $0$. $\endgroup$ – Parcly Taxel Dec 6 '17 at 1:47
  • $\begingroup$ I too have yielded 0, I am looking for proof. $\endgroup$ – Oiler Dec 6 '17 at 1:48
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Use $|\sin(x/n)|\leq|x/n|=x/n$ for $x>0$, $\left|\dfrac{n^{2}\sin(x/n)}{n^{3}x+x(1+x^{3})}\right|\leq\dfrac{nx}{n^{3}x+x(1+x^{3})}\stackrel{\sqrt{ab}\leq \frac{1}{2}(a+b)}{\leq}\dfrac{nx}{2(n^{3}x^{2}(1+x^{3}))^{1/2}}=\dfrac{1}{2n^{1/2}(1+x^{3})}\leq\dfrac{1}{2(1+x^{3})}$.

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    $\begingroup$ $a+b\geq 2\sqrt{ab}$ is used. $\endgroup$ – user284331 Dec 6 '17 at 1:49
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Without DCT: In absolute value the integrand is bounded above by

$$\frac{n^2 (x/n)}{n^3x + x(1+x^3)} = \frac{n}{n^3 + (1+x^3)} \le \frac{n}{n^3 + x^3}.$$

Let $x = ny$ to see that

$$\int_0^\infty \frac{n}{n^3 + x^3}\, dx = \frac{1}{n}\int_0^\infty \frac{1}{1 + y^3}\, dy.$$

Since the last integral converges, the integral in question is bounded above by a constant times $1/n,$ hence converges to $0.$

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  • $\begingroup$ Why the down vote? $\endgroup$ – zhw. Dec 6 '17 at 21:57
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$$\int\limits_0^\infty \frac{n^2 \sin(x/n)}{n^3x + x(1 + x^3)} \, d x=\int\limits_0^\infty \frac{n^3 \sin(y)}{n^4y + n y(1 +n^3 y^3)} \, d y=\frac{1}{n} \int\limits_0^\infty \frac{\sin(y)}{y + y^3 +y/n^3} \, d y$$ Here we make the change of variables $y=n x$. The integral $$ \int\limits_0^\infty \frac{\sin(y)}{y + y^4 +y/n^3}$$ is finite. Thus, the limit under discussion tends to zero.

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