Conclusion about limit definition of e^a for a sequence of real numbers {a_n} converging to a? I have seen this fact used in several demonstrations, but have never seen a proof of it.

I believe the statement is: If $\{a_n\}$ is a sequence of real numbers such that $a_n \rightarrow a$ finite, then $(1 + \frac{a_n}{n} )^n \rightarrow e^a $. 

Any help or references appreciated!
A simple approach starts from the fact that $\mathrm e^{x-x^2}\leqslant1+x\leqslant\mathrm e^x$ for every $x\geqslant-\frac12$ and that $\frac{a_n}n\geqslant-\frac12$ for every $n$ large enough, hence, applying this to $x=\frac{a_n}n$ for every $n$ such that $\frac{a_n}n\geqslant-\frac12$ yields
$$
\mathrm e^{nx-nx^2}\leqslant\left(1+\frac{a_n}n\right)^n\leqslant\mathrm e^{nx},
$$
that is,
$$
\mathrm e^{a_n-a_n^2/n}\leqslant\left(1+\frac{a_n}n\right)^n\leqslant\mathrm e^{a_n}.
$$
Since $a_n\to a$ and $\frac{a_n^2}n\to0$, the LHS and the RHS both converge to $\mathrm e^a$.
This is  called sequential continuity, one of the equivalent forms of continuity that applies to Real numbers , that says : $x_n\rightarrow x$ , then $f(x_n)\rightarrow f(x) $ , i.e., if the sequence $x_n$ converges to $x$, then the sequence $f(x_n)$ converges to $f(x)$. Note that there are spaces where continuity and sequential continuity are not equivalent. Can you tell how it applies here?

EDIT: More formally: if $a_n \rightarrow a$, we have,  $Lim_{n\rightarrow \infty} (1+a_n/n)^n=e^{Lim_n\rightarrow \infty a_n}=e^a$

EDIT 2: The argument above is correct, but incomplete. I will include a justification
when I can make it rigorous, or delete if I cannot do so in a couple of days.

There is a straightforward argument for why $(1+a_n/n)^n=e^{Lim_n\rightarrow \infty a_n}=e^a$: basically, $a_n$ will converge to $a$, and then $a_n$ will become a constant, so that the situation reduces to that of $Lim_{n\rightarrow \infty}(1+x/n)^n=e^x$:

For given $\epsilon&gt;0$ , we can find a positive integer $N$ with $|a_n-a|&lt; \epsilon$ for all $n&gt;N$ , and there is a positive integer $N'$ with $|(1+x/n)^n-e^x |&lt;\epsilon$ for all $n&gt;N'$. Now, choose $N_2&gt;$ $Max${${N,N'}$}. Then $$(1+(a-\epsilon)/N_2)^{N_2} &lt;(1+a/{N_2})^{N_2}&lt; (1+(a+\epsilon)/N_2)^{N_2}$$ Now let $N_2$ become even larger, so that $\epsilon$ gets increasingly-closer to $0$, and the left- and right- sides of the inequality will squeeze the middle term into being $e^a$