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: 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 $\mathrm e^a$.
A: No special estimation is needed. 
It is correct to conclude that $f_n(a_n)$ converges to $f(a)$ in this case, because for large enough $n$ (in fact for $n  > |a|$), all the $f_n$ are monotonic on an interval containing $a$.  The principle is

if $f_n(x) \to f(x)$ for all $x$, the limit $f(x)$ is continuous at $a$, and there is a $\delta > 0$ such that all $f_n$ are monotonic on $(a - \delta, a+ \delta)$ for $n \geq n_0$, then $f_n(a_n) \to f(a)$ for any sequence $a_n$ converging to $a$.

For a proof, take $\epsilon$ a positive parameter tending to $0$, $n$ large enough so that $a_n$ is in a small subinterval $(a- \epsilon,a+\epsilon)$ of the interval of monotonicity, then $f_n(a_n)$ is bounded between the min and max of $(f_n(a - \epsilon), f_n(a+\epsilon))$ and so is asymptotically bounded between $f(a - \epsilon)$ and $f(a + \epsilon)$. This sandwich of upper and lower bounds converges to $f(a)$ for small $\epsilon$.
Continuity of $f_n(x)$ was not used, only continuity at $x=a$ of the limit $f(x)$.
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 $\epsilon>0$ , 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 $e^a$
