$\lim_{n\to\infty} n^2 \int_{0}^{1} \frac{x\sin{x}}{1+(nx)^3} \, \mathrm{d}x$

Question

I have to calculate

$$ \lim_{n\to\infty} n^2 \int_{0}^{1} \frac{x\sin{x}}{1+(nx)^3} \, \mathrm{d}x $$

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I'm thinking of using the dominated convergence theorem. So I define

$$f_n(x)=\frac{n^2 x \sin{x}}{1+(nx)^3}.$$

$f_n$ are measurable owing to the fact that they are continuous. Now, I have to calculate its limit. If I'm not wrong:

$$\lim_{n\to\infty} \frac{n^2 x \sin{x}}{1+(nx)^3} = (x\sin{x}) \lim_{n\to\infty} \frac{n^2}{1+(nx)^3} = (x\sin{x}) \cdot 0 =0.$$

So $|f_n(x)|\le0$ and using the dominated convergence theorem:

\begin{align*} \lim_{n\to\infty} n^2 \int_{0}^{1} \frac{x\sin{x}}{1+(nx)^3} \, \mathrm{d}x &=\lim_{n\to\infty} \int_{0}^{1} n^2 \frac{x\sin{x}}{1+(nx)^3} \, \mathrm{d}x \\ &= \int_{0}^{1} \lim_{n\to\infty} n^2 \frac{x\sin{x}}{1+(nx)^3} \, \mathrm{d}x \\ &= \int_{0}^{1} 0 \, \mathrm{d}x =0 \end{align*}

Is it ok?

Answer 1

$|f_n(x)| \leq 0$ is certainly false. It is true that $f_n (x) \to 0$ for each $x$.

If $x\leq \frac 1 n$ then $|\frac {n^{2}x \sin x} {1+n^{3}x^{3}}|\leq n^{2}x \sin x\leq 1$ since $\sin x \leq x$. If $x >\frac 1n$ then $|\frac {n^{2}x \sin x} {1+n^{3}x^{3}}|\leq |\frac {n^{2}x \sin x} {n^{3}x^{3}}|\leq \frac 1 {nx}\leq 1$. Hence, $|f_n(x) | \leq 1$ in both cases. Hence, the limit of the integral is $0$ (by DCT).