This isn't a coursework problem. I'm on winter break. 1. The problem statement, all variables and given/known data A common approximation used in physics is: (1+x)n ≈ 1+nx for small x This implies that lim(x→0) (1+x)n = lim(x→0) 1+nx which is a true statement. However, lim(x→0) (1+x)n = lim(x→0) [(1+x)1/x]xn = lim(x→0) exn This leads to the fact that (1+x)n ≈ exn for small x The question is: which one is a better approximation, and how do I show it? 3. The attempt at a solution Compute the error terms. E1 = (1+x)n - (1+nx) E2 = (1+x)n - exn They intersect at E1 = E2, or at 1+nx = exn I'm not sure how to go from here. I'm pretty sure that x can't be isolated. However, since exn is always positive, I think the only root is at x=0. Wolfram gives a funny answer though. If anyone would like to explain Wolfram's answer, please do. http://www.wolframalpha.com/input/?i=1%2Bnx+%3D+e^%28nx%29 Assuming positive n, for a large x, exn is a worse approximation for the function. Since exn grows much faster than (1+nx), exn is always a worse approximation than (1+nx). Assuming negative n, for a large x, exn is a worse approximation for the function. Since exn doesn't grow as quickly as (1+nx) (it levels off), exn is always a better approximation than (1+nx). Or it may just be the case that 1+nx is always a better approximation, since it's more commonly used.