Exercise on a sequence of funcions $f_n\rightarrow f$ and Lebesgue Theorem. I have this sequence of functions for $n \ge 1$ as $n\rightarrow \infty$
$$a_n(x)=\frac{(\sin x)^n}{x^n}$$
For first I have to find $a_n\rightarrow x$ pointwise in $(0,+\infty)$ as $n\rightarrow \infty$.
My idea:
For $x\in (0,1)$ I have that $\sin x^n \approx x^n$ and $x^n \rightarrow 0$ when $n \rightarrow \infty$.
For $x>1$ I try to bound the sine:
$$-\frac{1}{x^n}\le\frac{(\sin x)^n}{x^n}\le\frac 1 {x^n}$$
Which clearly goes to $0$ as $n\rightarrow \infty$.
So I can say that $a_n \rightarrow x$ with $x=0$.

I want also compute $\lim_{n\rightarrow \infty}\int_0^{+\infty} a_n(x)\,dx.
$
Using the Lebesgue Dominated convergence theorem I have that $|a_n(x)|\le g(x)$ with $g(x)=0$.
So my integral is 
$$\int_0^{+\infty}|a_n(x)| \, dx=0$$
Am I right on my reasoning? Thank you.
 A: If $x\ne0$ then $|\sin x/x|<1$; then $|a_n(x)|=|\sin x/x|^n$ converges to $0$ as $n\to\infty$. If $x=0$, $a_n(0)=1$, Thus, $a_n$ converges pointwise to the function defined as $a(x)=0$ if $x\ne0$, $a(0)=1$.
now, to apply the dominated convergence theorem, you need an integrable function $g$ such that $|a_n(x)|\le g(x)$. To construct this $g$ we use two obvious inequalities: $|a_n(x)|\le1$ and $a_n(x)|\le1/|x|^n$. The first one will be used for small $x$, and the second one for large $x$. With this in mind, it is not too difficult to fall upon
$$
g(x)=\begin{cases}
1 &\text{if }|x|\le1,\\
\dfrac{1}{|x|^2} &\text{if }|x|>1.
\end{cases}
$$
A: Let $f_n(x):=(\sin(x)/x)^n$ for $n\geqslant 1$. For each $x>0$ we have $\sin(x)\leqslant x$ so
$$\frac{f_{n+1}(x)}{f_n(x)}=\frac{\sin(x)}{x}\leqslant 1$$
This implies that $(f_n(x))$ is a nonincreasing sequence for each $x>0$. Second we have
$$\int_0^{\infty}\frac{\sin(x)}{x}\,dx=\frac{\pi}{2}<\infty$$
and in general for $n\geqslant 2$: 
$$\Big|\int^{\infty}_0\Big(\frac{\sin(x)}{x}\Big)^n\,dx\Big|\leqslant \int^1_0\Big|\frac{\sin(x)}{x}\Big|^n\,dx+\int^{\infty}_1\Big|\frac{\sin(x)}{x}\Big|^n\,dx\\\leqslant\int^1_0\Big|\frac{\sin(x)}{x}\Big|^n\,dx+\int^{\infty}_1\frac{1}{x^n}\,dx<\infty$$
Hence $(f_n)$ is integrable on $(0,\infty)$. Moreover $f_n$ converges to the function $f$ with $f(x)=\sin(1)$ for $x=1$ and $f(x)=0$ else. Clearly $f$ is integrable with integral $\int f\,dx=0$. By monotone convergence theorem follows:
$$\lim_n\int^{\infty}_0f_n(x)\,dx=\int^{\infty}_0\lim_nf_n(x)\,dx=\int^{\infty}_0f(x)\,dx=0$$
