Show $\lim\limits_{\lambda\to\infty} e^{\lambda e^{\frac{t}{\sqrt{\lambda}}} - t\sqrt{\lambda} - \lambda} = e^{\frac{t^2}{2}}$.
In this case the underlying distribution is a Poisson Distribution with parameter $\lambda$.
Hint: Use Poisson Expansion.
Answer: Using the Taylor Series of $e^x$, I obtain:
$e^{\lambda e^{\frac{t}{\sqrt{\lambda}}} - t\sqrt{\lambda} - \lambda} = \sum\limits_{x = 0}^{\infty} \frac{\Big(\lambda e^{\frac{t}{\sqrt{\lambda}}} - t\sqrt{\lambda} - \lambda\Big)^x}{x!}$
However, I have no idea where to proceed from here.
Note that $$ - t\sqrt{\lambda} - \lambda + \lambda e^{\frac{t}{\sqrt{\lambda}}}= - t \lambda^{1/2} - \lambda + \lambda\sum_{n=0}^\infty \frac{t}{\lambda^{n/2}n!}\\ = \frac{t^2}{2!} + \frac{t^3}{\lambda^{1/2}3!} + \frac{t^4}{\lambda^{2/2}4!} + \cdots $$ It follows that $\lim_{\lambda \to \infty} {\lambda e^{\frac{t}{\sqrt{\lambda}}} - t\sqrt{\lambda} - \lambda} = \frac{t^2}{2}$.
Because the function $e^x$ is continuous, we can conclude that $\lim_{\lambda \to \infty} e^{\lambda e^{\frac{t}{\sqrt{\lambda}}} - t\sqrt{\lambda} - \lambda} = e^{t^2/2}$, which is what we wanted.