TITLE: A limit similar to the famous $\left(1 + \frac{1}{a_n}\right)^{a_n}$ one
QUESTION [6 upvotes]: Let's consider two sequences $(a_n), (b_n)$ in $\mathbb{R}$ such that 
$$\lim_{n \to +\infty} a_n, \lim_{n \to +\infty} b_n  = + \infty$$
Proposition: the sequence
$$x_n = \left(1 + \frac{1}{a_n}\right)^{b_n}$$
has a limit if $\lim_{n \to +\infty} \frac{b_n}{a_n}$ exists.
How can one prove it? What is this limit? Does $x_n$ have a limit if $\lim_{n \to +\infty} \frac{b_n}{a_n}$ does not exist?

REPLY [3 votes]: From the definition of real power
$$x_n = \left(1 + \frac{1}{a_n}\right)^{b_n} = \exp\left(b_n \ln \left(1 + \frac{1}{a_n}\right)\right) = \exp\left(\frac {b_n}{a_n} a_n \ln \left(1 + \frac{1}{a_n}\right)\right) = \exp\left(\frac {b_n}{a_n} \ln \left(1 + \frac{1}{a_n}\right)^{a_n}\right)$$
But the exponential function is continuous, so 
$$\lim e^{p_n} = e^r \iff \lim p_n = r$$
Notice that 
$$\lim \frac {b_n}{a_n} \ln \left(1 + \frac{1}{a_n}\right)^{a_n} = \lim \frac {b_n}{a_n}$$
which gives us the thesis