Let $b>0.$ Evaluate $\lim_{n \to \infty} \int^{b}_{0} \frac{\sin nx}{nx}dx$
Put $u = nx$. Then your integral becomes $$\int_0^{nb} \frac{1}{n} \frac{\sin u}{u} du = \frac{1}{n}\int_0^{nb} \frac{\sin u}{u} du .$$
Taking the limit as $n \to \infty$ and noting that if $\lim a_n = a$, $\lim b_n = b$ then $\lim a_nb_n = ab$ we have $$\int_0^b \frac{\sin x}{x} dx = \left(\lim_{n\to \infty} \frac{1}{n}\right) \times \frac{\pi}{2} = 0.$$