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 $|f_{n}|$?