Evaluate $\lim_{n \to \infty} \int_{1}^{2}\frac{\sin(nx)}{x}dx$

Integrate by parts, letting $u=\frac{1}{x}$ and $dv=\sin(nx)\,dx$. Then $du=-\frac{1}{x^2}\,dx$ and we can take $v=-\frac{\cos nx}{n}$.

Our integral is equal to $$\left. -\frac{1}{x}\cdot \frac{\cos(nx)}{n}\right|_1^2 -\int_1^2 \frac{\cos nx}{nx^2}\,dx.$$ Both parts $\to 0$ as $n\to\infty$.