I have to use the DCT in order to get the value of this limit:
$$ \lim_{n\to \infty} \int_0^n \biggl(\frac{\sin x}{x} \biggr)^n dx$$
If i take $f_n = \bigl(\frac{\sin x}{x} \bigr)^n, \lim_{n\to \infty} f_n =0 $ so it converges, but i can't tell what function to use for the domination.
My thought is that as $|\sin x| \le |x| $, and $\space \bigl|\left(\frac{\sin x}{x} \right)^n \bigr| \le \space \left|\frac{\sin x}{x} \right|^n $ then $ \space \left|\frac{\sin x}{x} \right|^n \le 1$ but i feel like I'm messing up hard somewhere.
How would you proof that function is dominated?
Thanks in advance.
To use DCT, you have to be on the same measure space. So you should view the original integral as $\int_\mathbb{R} (\frac{\sin x}{x})^n 1_{0 \le x \le n} dx$. Let $f_n(x) = (\frac{\sin x}{x})^n 1_{0 \le x \le n}$ and $f(x) = 1$ for $0 \le x \le 1$ and $f(x) = \frac{1}{x^2}$ for $x > 1$. Then $|f_n(x)| \le f(x)$ for each $n \ge 2$ and $x \ge 0$. Since $f \in L^1$, we can apply DCT.