As John suggested in the comments above: when x-->2 we handle a limit of type 0/0. In order to calculate it we use the derivative of the numerator and a derivative of the denominator:
f'(X^5-2^5) = 5x^4
---------- ----
f'(x-2) = 1
if we'll substitute x with 2 we'll get:
5*2^4
----- = 80
1