I am trying to figure out why this is true:
$ \lim_{p \to 0}\frac{1}{2p}\left((1+p)e^{-\frac{y}{1+p}} - (1-p)e^{-\frac{y}{1-p}}\right) = e^{-y} + ye^{-y}$
I have already tried L'Hopital's Rule, but it gave me something that I couldn't simplify. The problem seems to be the $\frac{1}{2p}$ term never seems to go away. I know the exponential function can be represented as: $e^x = \lim_{n\to\infty} (1+\frac{x}{n})^n$, but it doesn't seem immediately obvious how that would apply in this situation.
Thanks.