I know I'm not being explicit and I may be coinciding with somebody's preceding post (sorry about that if it is), but it seems to me to be the type of problem you can apply polar coordinates to to get a solution.
Here is another way using the Laplace transform.
$\displaystyle \displaystyle f(t)=\frac{\sin(t)}{t}=\sum_{n=0}^{\infty}\frac{(-1)^nt^{2n}}{(2n+1)!}$
Then via the laplace transform we get
$\displaystyle \displaystyle F(s)=\int_{0}^{\infty}\frac{\sin(t)}{t}e^{-st}dt=\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)!}\int_{0}^{\infty}t^{2n}e^{-st}dt$
Notice that $\displaystyle \displaystyle \lim_{s \to 0^{+}}F(s)=\int_{0}^{\infty}\frac{\sin(t)}{t}dt$
Now taking the transform above gives
$\displaystyle \displaystyle F(s)=\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)!}\left( \frac{(2n)!}{s^{2n+1}}\right)=\sum_{n=0}^{\infty}\ frac{(-1)^n}{(2n+1)}\left( \frac{1}{s^{2n+1}}\right)=\tan^{-1}\left( \frac{1}{s}\right)$
This gives $\displaystyle \displaystyle \lim_{s \to 0^+}\tan^{-1}\left( \frac{1}{s}\right)=\frac{\pi}{2}$