Calculus $L=\lim_{n\to \infty} \int_{0}^{\pi} \sqrt[n]{x}\sin x\: dx$ Calculus: $$L=\lim_{n\to \infty} \int_{0}^{\pi} \sqrt[n]{x}\sin x\: dx$$
My tried:
I think that 
We have $$\left|\sqrt[n]{x}\sin x\right|\leq 1$$
Hence
$$\lim_{n\to \infty} \int_{0}^{\pi} \sqrt[n]{x}\sin x\: dx=\int_{0}^{\pi}\sin x\: dx=2$$
 A: The dominated convergence theorem says that since there is a dominating function $g(x)= \pi$, that is $|x^{1/n} \sin x| \leq g(x)$ on $[0, \pi]$ for $n\geq 1$, and $\int_0^\pi g(x) dx < \infty$, we can pass the limit through:
$$\lim_{n \to \infty} \int_0^\pi x^{1/n} \sin x dx = \int_0^\pi (\lim_{n \to \infty} x^{1/n} \sin x) dx = \int_0^\pi \sin x dx = 2.$$
A: $$ \int_{0}^{\pi} \sqrt[n]{x}\sin x\: dx \leq \sqrt[n]{\pi} \int_{0}^{\pi}\sin x\: dx=2 \sqrt[n]{\pi} $$
$\forall \delta$, $0\lt \delta \lt \pi$
$$ \int_{0}^{\pi} \sqrt[n]{x}\sin x\: dx\geq   \int_{\delta}^{\pi}\sqrt[n]{x}\sin x\: dx\geq \sqrt[n]{\delta } \int_{\delta }^{\pi}\sin x\: dx=\sqrt[n]{\delta } (1+\cos \delta)$$
so
$$\sqrt[n]{\delta } (1+\cos \delta)\leq \int_{0}^{\pi} \sqrt[n]{x}\sin x\: dx\leq 2 \sqrt[n]{\pi}$$
Hence,
$$ 1+\cos \delta\leq\varliminf_{n\to \infty}\int_{0}^{\pi} \sqrt[n]{x}\sin x\: dx\leq \varlimsup_{n\to \infty}\int_{0}^{\pi} \sqrt[n]{x}\sin x\: dx\leq 2$$
Let $\delta\to 0$, we get 
$$\lim_{n\to \infty} \int_{0}^{\pi} \sqrt[n]{x}\sin x\: dx=2$$
