Existence of the limit $\lim_{R\to\infty} \int_0^R \frac{\sin x}{x}\; dx$

Question

I must prove the existence of the limit $$\lim_{R\to\infty} \int_0^R \frac{\sin x}{x}\; dx$$ The justification given is as follows: for all $R>\frac{\pi}{2}$ we have $$\begin{align}\int_0 ^R\frac{\sin x}{x}\; dx&=\int_0^{\pi/2}\frac{\sin x}{x}\; dx-\frac{\cos R}{R}-\int_{\pi/2}^R \frac{\cos x}{x^2}\; dx\\&\to \int_0^{\pi/2}\frac{\sin x}{x}\; dx -\int_{\pi/2}^\infty \frac{\cos x}{x^2}\; dx\end{align}$$ for $R\to \infty$ having used the Lebesgue Theorem.

1. Why does it have to be $R>\pi/2$?
2. Where was Lebesgue's theorem used?