sin(x)/x

when x=0, this is 0/0 which is undefined, but lim(x->0)(sin x)/x=1, as d(sin x)/dx = cos x = 1, dx/dx = 1, and 1/1=1. In fact, a(sin bx)/cx = ab/c as x approaches 0. It's provable.

As x approaches infinity, 1/x approaches 0, and so sin(x)*(1/x) approaches 0. But it's not so simple in other equations such as (e^x)/x where you have disparate limiting factors.

This function appears one of my textbooks (H Priestley, Introduction to integration) under the wonderfully descriptive heading:

Non-integrable functions which can be integrated

What Priestley's trying to say is that the function is not Lebesgue integrable, because the absolute value |sin(x)|/|x| is not integral, but the improper Riemann integral of sin(x)/x between 0 and infinity exists, and can be shown by some kind of analytic witchcraft ("an easy exercise") to equal pi/2.

So that's one for the hordes of you who thinks Lebesgue integration is the best thing since sliced bread. Priestley reckons it is a "small price" to pay, but I say Riemann integration gets the kudos for this one.