# How to show that $\lim_{p\to 0}\frac{pe^{2tp}}{1-e^{2tp}(1-p)}=(1+2t)^{-k}$

How would I find the following limit? I am thinking I might need to do a Taylor expansion of the exponentials but it hasn't worked out nicely. Thanks for the help.

$$\lim_{p\to 0} \left( \frac{pe^{2tp}}{1-e^{2tp}(1-p)} \right)^k$$

The answer is $(1+2t)^{-k}$.

-
If this is a homework exercise, you should add the homework tag. –  Brett Frankel Mar 7 '12 at 0:47
Naive application of L'Hopital's: $$\begin{eqnarray} \lim\limits_{p\to 0}\left(\frac{pe^{2tp}}{1-e^{2tp}(1-p)}\right)^k &=&\left(\lim\limits_{p\to 0}\frac{pe^{2tp}}{1-e^{2tp}(1-p)}\right)^k\\ &=&\left(\lim\limits_{p\to 0}\frac{e^{2tp}+2p^2e^{2tp}}{2pe^{2tp}(1-p)-e^{2tp}}\right)^k\\ &=&\left(\lim\limits_{p\to 0}\frac{e^{2tp}+2p^2e^{2tp}}{2pe^{2tp}-(e^{2tp}+2p^2e^{2tp})}\right)^k\\ \end{eqnarray}$$ Perhaps examine similarity to $\frac{x}{x-y}$? –  Alex Becker Mar 7 '12 at 0:57
Who is $k$ and why is (s)he butting in? –  André Nicolas Mar 7 '12 at 0:59
I think this is probability related and the $k$ powers the whole thing, but OP should confirm that. –  Patrick Da Silva Mar 7 '12 at 0:59
In my original post it was all to the power of k... not sure why it was changed –  icobes Mar 7 '12 at 1:00
Have you tried l'Hospital's rule? $$\lim_{p \to 0} \frac{pe^{2tp}}{1-e^{2tp}(1-p)} = \lim_{p \to 0} \frac{e^{2tp}+p(2t)e^{2tp}}{e^{2tp} - (1-p)(2t)e^{2tp}} = \lim_{p \to 0} \frac{1+2tp}{1-(1-p)(2t)} = \frac {1}{1-2t}$$ Just use the product rule and the case $k=1$ then.