Can anyone give me a hand on how to calculate the following using Lebesgue dominated convergence theorem:
$$\lim_{n\to\infty} \int_{[0,1]} \left(n\left(\sin\left(\frac xn\right)\right)\right)^n \mathsf dx$$
I wonder if as $n$ approaches $\infty$, $\left(n\left(\sin\left(\frac xn\right)\right)\right)^n$ becomes $0$ at every point in $[0,1)$ and becomes $1$ at $x=1$.
Any help would be greatly appreciated. Thank you.
For $x \in [0,1]$, $\sin(x/n) \leqslant x/n$, and
$$|f_n(x)| = \left[n \sin(x/n)\right]^n\leqslant x^n \leqslant 1.$$
Also,
$$ \lim_{n \to \infty}f_n(x) = \begin{cases} \lim_{n \to \infty}x^n=0, & \,\,0 \leqslant x < 1,\\ 1, & \,\,x= 1.\end{cases}$$
Now you can apply LDCT, since $|f_n(x)|$ is bounded on $[0,1]$.
You can show that $f_n(1) \to 1$ in a variety of ways, including L'Hospital's rule.
We have
$$\lim_{n \to \infty}\log f_n(1) = \lim_{n \to \infty}n \log \left[n\sin\left(\frac{1}{n}\right)\right]= \lim_{n \to \infty} \frac{\log \left[\frac{\sin\left(\frac{1}{n}\right)}{\frac{1}{n}}\right]}{\frac{1}{n}} \\= \lim_{x \to 0}\frac{\log\left(\frac{\sin x}{x}\right)}{x}.$$
Applying L'Hospital's rule,
$$\lim_{n \to \infty}\log f_n(1) = \lim_{x \to 0}\frac{\log\left(\frac{\sin x}{x}\right)}{x}= \lim_{x \to 0}\frac{x}{\sin x}\left(\frac{x \cos x -\sin x}{x^2}\right)\\=\lim_{x \to 0}\frac{x}{\sin x}\lim_{x \to 0}\frac{\cos x -x \sin x - \cos x}{2x}\\=\lim_{x \to 0}\frac{x}{\sin x}\lim_{x \to 0}\frac{-\sin x}{2}\\= 0.$$
Hence $\displaystyle \lim_{n \to \infty} f_n(1) = \exp[\lim_{n \to \infty} \log f_n(1)]= \exp(0) = 1.$