calculus - Show that $int^{infty}_{0}left(frac{sin(x)}{x}right)^2 < 2$

I`m trying to show that this integral is converges and $<2$
$$\int^{\infty}_{0}\left(\frac{\sin(x)}{x}\right)^2dx < 2$$
What I did is to show this expression:
$$\int^{1}_{0}\left(\frac{\sin(x)}{x}\right)^2dx + \int^{\infty}_{1}\left(\frac{\sin(x)}{x}\right)^2 dx$$
Second expression :
$$\int^{\infty}_{1}\left(\frac{\sin(x)}{x}\right)^2 dx < \int^{\infty}_{1}\left(\frac{1}{x^2}\right)^2dx = \lim\limits_{b\to 0} {-\frac{1}{x}}|^b_0 = 1 $$
Now for the first expression I need to find any explanation why its $<1$ and I will prove it.

I would like to get some advice for the first expression. thanks!

Answer

Hint: $$\lim_{x\to0}\frac{\sin x}{x}=1.$$