# Prove that $\lim_{n\to\infty} \int_{-1}^1 \frac{\sin(x^2+1)}{n^2 + x^3} dx = 0$

Prove that

$$\lim_{n\to\infty} \int_{-1}^1 \frac{\sin(x^2+1)}{n^2 + x^3} dx = 0$$The question before this was to state the fundamental theorem of calculus so maybe you'd have to use that when working out this.

• Hint: may be not. ${}{}{}$ – user99914 Jun 13 '16 at 9:14
• Use $|\frac{\sin(x^2+1)}{n^2+x^3}|\le \frac{1}{n^2-1}$ for $n>1$. – user90369 Jun 13 '16 at 9:19

You have that for $n \ge 2$
$$0 \le \left| \int \limits _{-1} ^1 \frac {\sin (x^2 + 1)} {n^2 + x^3} \ \Bbb d x \right| \le \int \limits _{-1} ^1 \left| \frac {\sin (x^2 + 1)} {n^2 + x^3} \right| \ \Bbb d x \le \\ \int \limits _{-1} ^1 \frac 1 {| n^2 + x^3 |} \ \Bbb d x \le \int \limits _{-1} ^1 \frac 1 {| n^2 - 1 |} \ \Bbb d x = \frac 1 {n^2 - 1} \int \limits _{-1} ^1 1 \ \Bbb d x = \frac 2 {n^2 - 1} \to 0$$
and now use the squeeze theorem (you have squeezed the modulus of your integral between $0$ and something that converges to $0$, so it itself must converge to $0$).
hint : use the fact that $$-1<\sin(x^2+1)<1$$ and use sandwich theorem (squeeze theorem )