For any $0<p<1$ and $r$ a positive integer, the probability function $$f(x)={{r+x-1}\choose{x}}p^r(1-p)^x \ \ \ \ \ \ x=0,1,2...$$ defines a random variable $X$.
I have computed the mgf of the above distribution to be $$m_X(u)=\Big(\frac{p}{1-(1-p)e^u)}\Big)^r \ \ \ \ \ \ \ u<\text{ln}((1-p)^{-1})$$
Define a new random variable by $\ Y=2pX\ $. Write down the mgf $m_Y(u)$ of $Y$ and show that $$\lim_{p\to 0} m_Y(u)=\Big(\frac{1}{1-2u}\Big)^r$$
I am unsure of how to solve this, but here is my attempt.
\begin{align*} m_Y(u)&=\mathbb{E}(e^{uY})\\ &=\mathbb{E}(e^{2pXu})\\ &=\sum_{x=0}^{\infty} e^{2pxu}{{r+x-1}\choose{x}}p^r(1-p)^x\\ &=\frac{p^r}{(1-(1-p)e^{2pu})^r} \sum_{x=0}^\infty \binom{r+x-1}{x} (1 - (1-p)e^{2pu})^r ((1-p)e^{2pu})^x \\ &=\frac{p^r}{(1-(1-p)e^{2pu})^r} \end{align*} But clearly in my response, $$\lim_{p\to 0} m_Y(u)\neq\Big(\frac{1}{1-2u}\Big)^r$$ I'm confident there's an error in my logic, but I don't know where.
L'Hoptal's rule works fine. But one can use teh taylor expansion as well as in, for small $p$ we can use the taylor expansion of $e^x$ and assume $2u \neq 1$, we get, $$ \frac{p}{1-(1-p)e^{2pu}} = \frac{p}{1-(1-p)(1+2pu + O(p^2))} = \frac{p}{1-1+p-2pu + 2up^2 + O(p^2)} $$ Summing things up and factor a $p$ leads to $$ \frac{p}{1-(1-p)e^{2pu}} =\frac{p}{p(1-2u+O(p))} = \frac{1}{1-2u+O(p)}\to\frac{1}{1-2u} $$ continuity of $f(z) = z^k$ now gives the result.