Mathematical Analysis advice Claim: Let $\delta>0, n\in N.  $ Then $\lim_{n\rightarrow\infty} I_{n} $exists, where $ I_{n}=\int_{0}^{\delta} \frac{\sin\ nx}{x}  dx  $
Proof: $f(x) =\frac{\sin\ nx}{x}$ has a removable discontinuity at $x=0$ and so we let $f(0) =n$
$x = \frac{t}{n}$ is continuous and monotone on $t\in[0,n\delta]$, hence, $ I_{n}=\int_{0}^{n\delta} \frac{\sin\ t}{t}  dt  $ 
For all  $p\geq 1$, given any $\epsilon \gneq 0,  \exists n_{0} \gneq \frac{\epsilon}{2\delta}$ such that $\forall n\gneq n_{0}$,   $|I_{n+p}-I_{n}| =  \left\lvert\int_{n\delta}^{(n+p)\delta} \frac{\sin\ t}{t}  dt \right\lvert  \leq \int_{n\delta}^{(n+p)\delta} \frac{\left\lvert\sin t\ \right\lvert}{t}  dt = \frac{1} {n\delta}\int_{n\delta}^{M} \left\lvert\sin t\right\lvert dt \ \leq \frac{2} {n\delta} \lneq \epsilon$ ,  for some $M \in [n\delta, (n+p)\delta]$, by Bonnet's  theorem
Is my proof valid? Thank you. 

I can suggest a alternative path. Prove that $$\lim_{a\to 0^+}\lim_{b \to\infty}\int_a^b\frac{\sin x}xdx$$
exists as follows: integrating by parts
$$\int_a^b \frac{\sin x}xdx=\left.\frac{1-\cos x}x\right|_a^b-\int_a^b\frac{1-\cos x}{x^2}dx$$
Then use $$\frac{1-\cos h}h\stackrel{h\to 0}\to 0$$ $$\frac{1-\cos h}{h^2}\stackrel{h\to 0}\to\frac 1 2$$ $$\int_1^\infty\frac{dt}{t^2}=1<+\infty$$
Then it will follow your limit is indeed that integral, since changing variables and since $\delta >0$ $$\int_0^{n\delta}\frac{\sin x}xdx\to\int_0^\infty\frac{\sin x}xdx$$
Then it remains to find what that improper Riemann integral equals to. One can prove it equals $\dfrac \pi2$. First, one uses that 
$$1 +2\sum_{k=1}^n\cos kx=\frac{\sin \left(n+\frac 1 2\right)x}{\sin\frac x  2}$$
from where it follows $$\int_0^\pi\frac{\sin \left(n+\frac 1 2\right)x}{\sin\frac x  2}dx=\pi$$ since all the cosine integrals vanish. Now, the function $$\frac{2}{t}-\frac{1}{\sin\frac t 2}$$ is continuous on $[0,\pi]$. Thus, by the Riemann Lebesgue Lemma, $$\mathop {\lim }\limits_{n \to \infty } \int_0^\pi  {\sin \left( {n + \frac{1}{2}} \right)x\left( {\frac{2}{x} - {{\left( {\sin \frac{x}{2}} \right)}^{ - 1}}} \right)dx}  = 0$$
It follows that $$\mathop {\lim }\limits_{n \to \infty } \int_0^\pi  {\frac{{\sin \left( {n + \frac{1}{2}} \right)x}}{x}dx}  = \frac{\pi }{2}$$ so $$\mathop {\lim }\limits_{n \to \infty } \int_0^{\pi \left( {n + \frac{1}{2}} \right)} {\frac{{\sin x}}{x}dx}  = \frac{\pi }{2}$$
Since we know the integral already exists, we conclude $$\int_0^\infty  {\frac{{\sin x}}{x}dx}  = \frac{\pi }{2}$$
