Given $x_{n} \to x_{0}$ as $n \to \infty$, and $e^{x}=\sum_{k=0}^{\infty}\frac{x^{k}}{k!}$, prove that $\lim_{n \to \infty}e^{x_{n}} = e^{x_{0}}$ Problem: Given a convergent sequence $x_{n} \to x_{0}$ as $n \to \infty$, and that e is defined as $e^{x}=\sum_{k=0}^{\infty}\frac{x^{k}}{k!}$, prove that $\lim_{n \to \infty}e^{x_{n}} = e^{x_{0}}$.  
Now I know I could simply use limit rules, and say that $\lim_{n \to \infty}e^{x_{n}} = e^{\lim_{n \to \infty}x_{n}} = e^{x_{0}}$. However I would like (and the question suggests) to use that infinite series definition of $e^{x}$ to arrive at the answer.  
So far I have $\lim_{n \to \infty}e^{x_{n}} = \lim_{n \to \infty}\sum_{k=0}^{\infty}\frac{x_{n}^k}{k!} =\sum_{k=0}^{\infty}\frac{x_{0}^k}{k!}$, but again without just distributing the limit inside the summation,  I am stuck as to how to proceed from there and find the limit of that series. Any help would be appreciated! (It could be there is no way to do it without simply distributing the limit, I'm not sure)

Suppose we can prove that $x_k\to 0\Rightarrow e^{x_k}\to 1.$  Then, if $x_k\to x_0,\ y_k:=x_k-x_0\to 0$ and then $e^{y_k}=e^{x_k-x_0}\to 1$ and this implies that $e^{x_k}\to e^{x_0}$. 
So, it suffices to prove the result for $x_0= 0.$ But this is easy: choose $K$ large enough so that $k>K\Rightarrow |x_k|<\epsilon<1.$ Then, for such $k$
$ \left|\sum_{k=0}^\infty \frac{x_n^k}{k!} - 1\right|= \left|\sum_{k=1}^\infty \frac{x_n^k}{k!}\right|\le \sum_{k=1}^\infty \frac{\epsilon^k}{k!}\le\sum_{k=1}^\infty \epsilon^k=\frac{1}{1-\epsilon}-1$
The result follows (since $\epsilon$ is arbitrary).
