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 $\epsilon>0$ , we can find a positive integer $N$ with $|a_n-a|< \epsilon$ for all $n>N$ , and there is a positive integer $N'$ with $|(1+x/n)^n-e^x |<\epsilon$ for all $n>N'$. Now, choose $N_2>$ $Max${${N,N'}$}. Then $$(1+(a-\epsilon)/N_2)^{N_2} <(1+a/{N_2})^{N_2}< (1+(a+\epsilon)/N_2)^{N_2}$$ 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 $e^a$