General formula of $\lim_{n \to \infty} (1+a_n)^{b_n}$ where $b_n \to \infty$ 
What is the general formula of $$\lim_{n \to \infty} (1+a_n)^{b_n} \quad \text{where} \quad \lim_{n \to \infty} b_n = \infty?$$

For example we have $$\lim_{n\to \infty} \left(1 + \frac1n\right)^n = e$$
Now I want a general formula for $\lim_{n \to \infty} (1+a_n)^{b_n}$. I know if $a_n \to A \neq 0$ then the answer is infinity. But I am not sure about what will happen when $a_n \to 0$. Maybe the answer is $e^c$ where $c = \lim _{n \to \infty} b_n a_n$. But even if this is the answer, I want the proof.
It seems it must be a famous problem but unfortunately, I couldn't find the answer in math.SE. If an answer exists with proof, please post its link in the comments and I will delete my question.
 A: If the limit does exist, we take logs of both sides, getting
$$
\lim_{n \to \infty} (1+a_n)^{b_n}
= \lim_{n \to \infty} \exp\left(b_n \ln (1+a_n)\right)
= \exp \left( \lim_{n \to \infty} b_n \ln (1+a_n)\right)
= \exp \left( \lim_{n \to \infty} \frac{\ln (1+a_n)}{1/b_n}\right)
$$
So if $a_n \to 0$, we end up with a $0/0$ form, and L'Hospital's rule applies, and the rest depends on the particular dependencies of $a_n$ and $b_n$ on $n$....
A: Let $c_n := a_n b_n$ and assume that $0\lt a_n\rightarrow 0$, $b_n\rightarrow \infty$. Then define:
$$
A_n:={\left(1+a_n\right)^{b_n}}=\left({\left(1+a_n\right)^{a_n^{-1}}}\right)^{c_n}
$$
From this we can clearly observe that the limit of $A_n$ indeed depends on $c_n$. In particular, if $c_n \rightarrow B$, for some $B$, then $A_n\rightarrow e^{B}$. Otherwise, if $c_n$ is divergent, then so is $A_n$.
A: Let $b_n,a_n$ such that $\lim_n a_nb_n=c$. Moreover assume that $a_n\rightarrow 0$.
Then we have 
$$(1+a_n)^{b_n} = (1+a_n)^{\frac{c}{a_n}}  \cdot (1+a_n)^{b_n-\frac{c}{a_n}}$$
We deal with each term alone:
First, since $(1+\frac{1}{n})^n\rightarrow e$ we have that $(1+\frac{1}{n})^{cn}\rightarrow e^c$. As $a_n\rightarrow 0$ we can substitute $a_n$ with $\frac{1}{n}$.
It is therefore enough to show that $(1+a_n)^{b_n-\frac{c}{a_n}}\rightarrow 0$.
For every $\varepsilon>0$ there exists $N$ such that for all $n>N$ we have $|a_nb_n-c|<\varepsilon$. This implies that $|b_n-\frac{c}{a_n}|<\frac{\varepsilon}{a_n}$.
Therefore $(1+a_n)^{b_n-\frac{c}{a_n}} < (1+a_n)^{\frac{\varepsilon}{a_n}}$ and the latter convergence to $e^\varepsilon$. As $\varepsilon$ is arbitrary small, $e^\varepsilon$ is arbitrary close to $1$ and this completes the proof.
