I need to show that this lim goes to a function of e, namely $e^\frac{t^2}{2}$. I'm not sure if l'hospital's rule is needed. This is involved in the proof that the gamma distribution goes to the normal distribution (if anyone is interested). thanks!
$$ \lim \limits_{x \to \infty}{e^{-t{\sqrt x}}\left(1-\frac{t}{\sqrt x}\right)^{-x} }$$
The first-order approximation here is that $\left(1-\frac{t}{\sqrt{x}}\right)^{-x}\approx e^{t\sqrt{x}}$. Multiply that by $e^{-t\sqrt{x}}$, and we just get $1$. Clearly, this wasn't precise enough for our purposes.
All right, we need a better approximation like that. First, products are annoying to deal with, so let's take logarithms; if $L$ is the limit we seek, $$\begin{align} \ln L & = \ln \lim_{x\to\infty} e^{-t\sqrt{x}}\left(1-\frac{t}{\sqrt{x}}\right)^{-x} \\& = \lim_{x\to\infty} \ln \left ( e^{-t\sqrt{x}}\left(1-\frac{t}{\sqrt{x}}\right)^{-x} \right ) \\ & = \lim_{x\to\infty} -t\sqrt{x} -x\ln\left(1-\frac{t}{\sqrt{x}}\right) \end{align}$$ Next, let's get rid of those square roots; substitute $y=\sqrt{x}$: $$\ln L = \lim_{y\to\infty} -ty - y^2\ln\left(1-\frac ty\right)$$ From here - we could do something with L'Hopital's rule, but I greatly prefer power series as a tool, especially since we need more than just the first term. \begin{align*}\ln\left(1-\frac ty\right) &= -\frac ty - \frac12\cdot\frac{t^2}{y^2} - \frac13\cdot \frac{t^3}{y^3}-\cdots\\ \ln L &= \lim_{y\to\infty} -ty -y^2\left(-\frac ty - \frac12\cdot\frac{t^2}{y^2} - \frac13\cdot \frac{t^3}{y^3}-\cdots\right)\\ \ln L &= \lim_{y\to\infty} -ty +ty + \frac12\cdot t^2 + \frac13\cdot\frac{t^3}{y}+\cdots\\ \ln L &= \lim_{y\to\infty} \frac12\cdot t^2 + \frac13\cdot\frac{t^3}{y}+\cdots = \frac12t^2\\ L &= e^{\frac12t^2}\end{align*}