Use Dominated Convergence Theorem to prove that $\lim_{n\rightarrow \infty}\int_0^\pi\sqrt{\frac{t}{n}}\sin(\sqrt{\frac{n}{t}})\,dt$ Here it is function,
$$\lim_{n\rightarrow \infty}\int\limits_{0}^{\pi}\sqrt{\frac{t}{n}}\sin\sqrt{\frac{n}{t}}\;dt$$
but I read about Dominated convergence theorem and I don't know how to implement it to provie that that limit equals zero?
 A: Step A. Set $\,\displaystyle f_n(t)=\sqrt{\frac{t}{n}}\sin\sqrt{\frac{n}{t}},$ and
observe that
$$
\lvert\, f_n(t)\rvert=
\left|\sqrt{\frac{t}{n}}\sin\sqrt{\frac{n}{t}}\,\right|\le 1=g(t),$$
for all $t\in[0,\pi]$ and $n\in\mathbb N$, since
$$
\lvert \sin x\rvert\le \lvert x\rvert,
$$
for all $x$, and clearly $\int_0^\pi g(t)\,dt=\pi<\infty$.
Step B. Next, we have that 
$$\lim_{n\to\infty}f_n(t)=\lim_{n\to\infty}\frac{\sin\sqrt{\frac{n}{t}}}{\sqrt{\frac{n}{t}}}=0,
$$
for all $t\in [0,\pi]$.
Thus, Lebesgue Dominated Convergence Theorem is applicable and provides that 
$$
\lim_{n\to\infty}\int_0^\pi f_n(t)\,dt=\int_0^\pi\big(\lim_{n\to\infty}f_n(t)\big)\,dt=0.
$$
A: Hint:
a) Find an integrable function  $g$ on the given domain such that $|\sqrt{\frac{t}{n}}\sin(\sqrt{\frac{t}{n}})|$ $\leq g$ for all $n\in N$ almost everywhere.
b) Exchange limit and integral operations
A: No need for heavy machinery here. Note that
$$\left|\int\limits_0^\pi\sqrt\frac tn\sin\sqrt\frac nt\;dt\right| \le \int_0^{\pi} \left| \sqrt\frac tn\sin\sqrt\frac nt \right|\,dt$$
Since $|\sin(x)| \leq 1$ for all $x$ this is at most
$$\int_0^{\pi} \sqrt\frac tn\,dt$$
$$= {1 \over \sqrt{n}} \int_0^{\pi} \sqrt{t}\,dt$$
You don't even have to do the integral here; take limits as $n$ goes to $\infty$ and get 0.
A: This doesn't use the Dominated Convergence Theorem, but suppose we take
$$F(n) = \int\limits_0^\pi \sqrt{\frac{t}{n}}\sin\sqrt{\frac{n}{t}}\,dt$$
By the intermediate value theorem for integrals, 
$$F(n) = \pi\sqrt{\frac{\tau}{n}}\sin\sqrt{\frac{\tau}{n}}$$
for some $\tau \in (0, \pi)$.  
It follows that $|F(n)| < \frac{\pi^{\frac{3}{2}}}{n^{\frac{1}{2}}}$. (In fact, since $\sin$ is nonnegative over the domain of integration, $0 \le F(n)$, but we don't need this tighter bound). From here, the proof that $F(n) \to 0$ an $n \to +\infty$ follows from a quick application of the squeeze theorem.
A: I don't know about the DCT, but what about
$$\left|\int\limits_0^\pi\sqrt\frac tn\sin\sqrt\frac nt\;dt\right|\le\sqrt\frac1 n \int\limits_0^\pi \sqrt t\;dt=\frac2{3\sqrt n}\pi\sqrt\pi\xrightarrow[n\to\infty]{}0\;\;\;?$$
