Let m<n. We want to show that as , where . But and coincide on the intervals [0,1-(1/m)] and [1,2]. So we need only worry about the integral on the interval from 1-(1/m) to 1. Even on this interval, the integral is a bit messy to evaluate exactly, so I suggest using a little trick.
It's easy to see that if m<n then . Therefore , and the integral that we want to evaluate is less than . That integral is (very!) easy to evaluate, and it gets small as m gets large.