Proving that this integral is divergent

Question

$$\int_{0}^{\infty}\frac{|\sin x|}{x^2}$$

I have thought a lot but the absolute value just messes every way I try.

Answer 1

$$\lim_{x\to 0^+}\left(\frac{|\sin x|}{x^2}:\frac{1}{x}\right)=1\ne 0\;\wedge\;\int_0^1\frac{dx}{x}\text{ divergent }$$ $$\Rightarrow \int_0^1\frac{|\sin x|}{x^2}\text{ divergent }\Rightarrow \int_0^{+\infty}\frac{|\sin x|}{x^2}\text{ divergent.}$$