In a Measure Theory past exam, we are asked to prove that this limit is true.
$\lim_{n \to \infty}\int_0^1\frac{\sqrt{n^3}\sin x}{1 + n^2 x^2}d\lambda(x) = 0$
I have one path to a solution:
$$\int_0^1\frac{\sqrt{n^3}\sin x}{1 + n^2 x^2}d\lambda(x) \leq \int_0^1 \frac{1}{2n^{1/2}}\frac{2x n^2}{1 + n^2 x^2}d\lambda(x) = \frac{1}{2n^{1/2}}\log(1 + n^2).$$
Now since $1 \leq \lim_{n \to \infty} (1+n^2)^{n^{-1/2}} = \lim_{n \to \infty} (1+n^4)^{1/n} \leq \lim_{n \to \infty} ((1+\epsilon)^n)^{1/n} = 1 + \epsilon $ holds $\forall \epsilon > 0$, we can see the integral tends to $0$
But clearly what I have given here is not a "Measure Theory answer", and I also don't think it's a very good solution.
In my attempt for a more measure theoretical answer I did try to uniformly bound the integrand in norm by some integrable function so that I could apply the Dominated Convergence Theorem, but I failed at that; in particular it seems hard because the integrands (when plotted) grow very large, reaching their maxima closer and closer to $0$.
Hint to apply the Dominated Convergence Theorem. Note that for $x\in(0,1]$: $$0\leq \frac{\sqrt{n^3}\sin x}{1 + n^2 x^2}\leq \frac{\sqrt{n^3}x}{1 + n^2 x^2}\leq \frac{1}{\sqrt{x}}$$ which hold because, by letting $\sqrt{nx}=t>0$, $$t^3\leq \max(1,t^4)\leq 1+t^4.$$