Evaluate $\lim_{n \rightarrow \infty} \int_{0}^{\pi} \frac{\sin(nx)}{nx}dx$. Explanation of certain steps in procedure. I've been asked to evlauate $$\lim_{n \rightarrow \infty} \int_{0}^{\pi} \frac{\sin(nx)}{nx} dx$$
I have a few questions about particular things with the proof, I'm going to write them in bold to differentiate them from the rest of the text.
So we take $\epsilon > 0$.
We split the integral up as such: $$\lim_{n \rightarrow \infty} \int_{0}^{\pi} \frac{\sin(nx)}{nx} dx = \lim_{n \rightarrow \infty} \Bigg(\int_{0}^{\epsilon} \frac{\sin(nx)}{nx} dx +  \int_{\epsilon}^{\pi} \frac{\sin(nx)}{nx} dx \Bigg)$$
Dealing with $ \int_{\epsilon}^{\pi} \frac{\sin(nx)}{nx} dx$ first:
We have to examine $\frac{\sin(nx)}{nx}$. For $x \in [\epsilon, \pi]$ we have:
$$\Bigg|\frac{\sin(nx)}{nx} - 0\Bigg| \leq \frac{1}{n \epsilon}$$
Question 1: Why are we trying to establish this bound? Is it because we want to show that $\frac{\sin(nx)}{nx}$ converges uniformly to $0$ and as such be able to apply integral or dominating convergence? if that is the case how did we get zero? Because I tried using L'Hôpital to get a limit and I ended up with a value of $1$ if we are differentiating with respect to $x$, the same would occur if I differentiated with respect to $n$.
From  here the next step would be:
$$\lim_{n \rightarrow \infty}\int_{\epsilon}^{\pi} \frac{\sin(nx)}{nx} dx \leq \lim_{n \rightarrow \infty}  \int_{\epsilon}^{\pi}\frac{1}{n \epsilon}  = \lim_{n \rightarrow \infty} \frac{\pi - \epsilon}{n \epsilon} = 0$$
Question 2: We established that the upper bound went to 0. I take it this implies that our integral goes to zero as well? Is this what we have to accept because it is extremely difficult to compute our integral explicitly?
From here a similar procedure would occur with the other portion of the integral. The same questions I have about these two steps appear there.
Question 3: What is the purpose of this procedure? Is it because this is a hard integral to compute explicitly and as such we have to find approximate ways to establish what the integral is? Why do we take the limit towards a troublesome endpoint of the inner function ?
Note: I'm aware that there is a solution to this problem in another post, but I wanted to get clarity on some of the intricacies that are spinning my head.
Here is the original solution: Prove that $\lim_{n\rightarrow \infty} \int_{0 }^{\pi} \frac{\sin(nx)}{nx}dx=0$
 A: 1) We're just trying to get a bound on the integrand, and hence on the integral, exactly as you showed in your next step.
2) The fact that $\frac{\sin(nx)}{nx}\ge -\frac{1}{n\epsilon}$ gives you a lower bound of zero.
3) Yes, the purpose is to use other techniques in lieu of a closed form solution for the integral. Indeed, this integral is a non-elementary special function, the sine-integral function, and although we could just look up the asymptotic behavior of this (which is effectively what is being done in the solution that transforms to $\frac{1}{n}\int_0^{\pi n}\frac{\sin(x)}{x}dx$ and then notes that $\int_0^\infty \frac{\sin x}{x}dx=\frac{\pi}{2}$) this somewhat misses the point. 
The reason we cut off $\epsilon>0$ is actually not due to any irregularity with the integrand there: the integrand is completely regular at the origin. It is actually just so our bounding function has the nice form $\frac{1}{n\epsilon}$ that decays with $n$ whereas $\frac{\sin(nx)}{nx}$ always goes to one at the origin. Really we could just bound it with the piecewise function that is $1$ below $\epsilon$ and $\frac{1}{n\epsilon},$ which is exactly what's being done when the integral gets split up and the lower piece is seen to be bounded by $\epsilon$. 
(However, note that it is in fact often a good idea to split up the integral when one or both of the endpoints have singular behavior in the integrand.)

For clarity, here is a paraphrase of the solution: For any given $\epsilon>0,$ we have $\left|\frac{\sin(nx)}{nx}\right|\le 1$ for $x\le \epsilon$ and $\left|\frac{\sin(nx)}{nx}\right|\le \frac{1}{n\epsilon}$ for $x\ge \epsilon.$ Hence $$ \left|\int_0^\pi \frac{\sin(nx)}{nx}dx\right|\le \int_0^\pi\left|\frac{\sin(nx)}{nx}\right|dx\le \int_0^\epsilon dx + \int_\epsilon^\pi\frac{1}{n\epsilon}dx = \epsilon + \frac{\pi-\epsilon}{n\epsilon}.$$ So we have $$\lim_{n\to\infty} \left|\int_0^\pi \frac{\sin(nx)}{nx}dx\right|\le \epsilon.$$ Since $\epsilon$ was arbitrary, the limit must be zero.
A: $|\frac {\sin nx}{nx}| \le 1.$ This is true becuase $|\sin x| \le $x$ for all $x$
and $|\frac {\sin nx}{nx}| \le \frac {1}{nx}$ as $|\sin nx| \le 1$
$0\le |\int_0^{\pi} \frac {\sin nx}{nx}\ dx| \le \int_0^{\pi} |\frac {\sin nx}{nx}|\ dx$
$\int_0^{\pi} |\frac {\sin nx}{nx}|\ dx \le \int_0^{\frac{1}{n}} 1\ dx + \int_{\frac{1}{n}}^\pi \frac {1}{nx}\ dx$
$\frac {1}{n} + \frac{1}{n}\ln \pi - \frac 1n ln \frac 1n$
Taking the limit as $n$ goes to infinity, we get 0.
