Q:

How to prove the limit of "the exponential of a sequence"

So given a convergent sequence $\{a_n\}_{n=1}^\infty$ with limit $a$, I'd like to prove that
$$\lim_{n\to\infty} \left(1+\frac{a_n}{n}\right)^n=e^a.\quad(1)$$
Knowing that $e$ is defined by
$$e=\lim_{n\to\infty} \left(1+\frac{1}{n}\right)^n,$$
the relationship in $(1)$ certainly not unintuitive, and also very useful, but how do I prove it?
In the case of a constant sequence $a_n=k\;\forall n$, it's pretty straightforward as you can write
$$\left(1+\frac{k}{n}\right)^n=\exp\left[\log\left(\left(1+\frac{k}{n}\right)^n\right)\right]=\exp\left[n\log\left(1+\frac{k}{n}\right)\right]=\exp\left[\frac{\log\left(1+\frac{k}{n}\right)}{1/n}\right]$$
and then taking the limit you can apply L'Hôpital's rule to differentiate the numerator and denominator separately and then get the result after a few manipulations. 
But since 
$$\frac{\mathrm{d}}{\mathrm{d}x} \log\left(1+\frac{f(x)}{x}\right)=\frac{xf'(x)-f(x)}{xf(x)+x²}$$
I will need to know the derivative $(a_n)'$ with respect to $n$ of the sequence, to use this approach, which is not necessarily well-defined. Is there another way to go about this?

A:

Nothing new here, but perhaps streamlined a bit: Apply $\ln$ to get $$(1)\,\,\,\,n\ln (1+a_n/n)=\frac{\ln (1+a_n/n)}{a_n/n}\cdot a_n.$$ Now as $h\to 0,$ $\ln(1+h)/h = (\ln(1+h)-\ln 1)/h \to \ln'(1) = 1\,$ by definition of the derivative. It follows that the limit in (1) is $a.$ Exponentiating back gives the limit of $e^a$ as desired.  

