If $a>0$, show $\text{lim}_{n \rightarrow \infty} \int_{a}^{\pi}\frac{sin(nx)}{nx}dx = 0$. What happens if $a=0$? If $a>0$, show $\text{lim}_{n \rightarrow \infty} \int_{a}^{\pi}\frac{sin(nx)}{nx}dx = 0$. What happens if $a=0$?
Should I use a delta-epsilon proof here or is there a better way?
 A: \begin{align*}
\int_{0}^{\pi}\dfrac{\sin(nx)}{nx}dx=\dfrac{1}{n}\int_{0}^{n\pi}\dfrac{\sin x}{x}dx\rightarrow 0\cdot\dfrac{\pi}{2}=0.
\end{align*}
A: It doesn't matter whether $a>0$ or $a=0$. Make the substitution $y=nx$ to get $\frac  1 n \int_{na}^{n\pi} \frac {sin y} y  dy \to 0$ since $\int_0^{\infty} \frac {sin y} y  dy$ is finite ($=\pi /2)$. 
$\int_c^{d}\frac {sin y} y  dy$ is bounded when $c,d$ range over $[0,\infty)$. 
A: If you have never seen the evaluation of the integral
$\int_0^\infty \frac {\sin x}{x} \ dx$ you can bypass evaluating that integral, but you will need to be a little bit creative.
$0\le|\int_a^\pi \frac {\sin nx}{nx} dx| \le \int_a^\pi |\frac {\sin nx}{nx}| dx \le \int_a^\pi \frac {1}{nx} dx = \frac 1n \ln(\frac {\pi}{a})$
And by the squeeze theorem this will the limit will equal $0$ as $n$ goes to infinity, and $a>0$
What about when $a = 0$?  We need to make a small tweak.
We will use $|\sin nx| \le |nx| dx$
$\int_0^\pi \frac {\sin nx}{nx} dx = \int_0^\frac {\pi}{2n} \frac {\sin nx}{nx} dx + \int_\frac {\pi}{2n}^\pi \frac {\sin nx}{nx} dx$
$\int_0^\frac {\pi}{2n} \frac {\sin nx}{nx} dx \le \frac {\pi}{2n}$
$|\int_0^\pi \frac {\sin nx}{nx} dx| \le \frac {\pi}{2n} + \frac 1n \ln 2n$
