What is the distribution of the m.g.f. that corresponds to the limit. Let $\lambda>0$ be a fixed number. Suppose $X_n\sim Bin(n,\frac{\lambda}{n})$, for $n\in\mathbb{N}\setminus\{0\}$.

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*Find the moment-generating function $M_{X_n}(t)$ for $X_n$.

*Compute for $t$ fixed, $\lim_{n\rightarrow\infty}M_{X_n}(t)$. What is the distribution of the m.g.f. that corresponds to the limit?

My Attempt

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*$\displaystyle{M(t)=\mathbb{E}[e^{tx}]=\sum_{x=0}^xe^{xt}\binom{n}{x}\left(\frac{\lambda}{n}\right)^x\left(1-\frac{\lambda}{n}\right)^{n-x}=\sum_{x=0}^x\binom{n}{x}\left(\frac{\lambda e^{t}}{n}\right)^x\left(1-\frac{\lambda}{n}\right)^{n-x}=\left(\frac{\lambda e^t}{n}+1-\frac{\lambda}{n}\right)^n}$


*\begin{align}
\lim_{n\rightarrow\infty}\left(\frac{\lambda e^t}{n}+1-\frac{\lambda}{n}\right)^n&=\lim_{n\rightarrow\infty}\left(1+\frac{\lambda e^t-\lambda}{n}\right)^n\\
&=\lim_{n\rightarrow\infty}e^{n\ln(1+\frac{\lambda e^t-\lambda}{n})}\\
&=\lim_{n\rightarrow\infty}(\lambda e^t-\lambda)+\frac{o(n)}{n}\quad\text{(where $\lim_{n\rightarrow\infty}\frac{o(n)}{n}=0$)}\\
&=\lambda e^t-\lambda
\end{align}
The part that I am having trouble with is figuring out which distribution the limit of the m.g.f. corresponds to.

You missed an expoenential at the end. The correct limit is $e^{\lambda (e^{t}-1)}$ and this is the MGF of Poisson distribution with parameter $\lambda$.
