TITLE: "Boundary" of convergence of  $\frac{1-(1-c^{n})^{2n}}{(1-c)^{2n}}$
QUESTION [6 upvotes]: I ran across this confounding limit I am wondering about.  It is as follows:
$$\displaystyle \lim_{n\to \infty}\frac{1-(1-c^{n})^{2n}}{(1-c)^{2n}}, \;\ 0<c<1$$
I played around with this on Maple and found that if c is less than approximately .382 (but 
greater than 0), it converges to 0. If c is greater than .382 (but less than 1), it 
diverges. What is it about .382?.
.382 is an approximation. By playing around more I could have taken it out to more decimal places.
The actual problem asks to prove that the above limit is < $\frac{1}{p(n)}$, where p(n) is a polynomial. 
I was mainly wondering how to solve the limit and why .382 is so significant.
Thank you all very much. You are always a big help.

REPLY [3 votes]: My previous comment was wrong of course. One can solve the problem using L'Hopital's rule. $0.38196$ appears as the solution to the equation $x^2-3x+1=0$ and equals $\frac{3-\sqrt{5}}{2}$ to be precise.