Given two sequences $a_n \to a$, $b_n \to b$, does $\lim\limits_{n \to \infty} a_n^{b_n} = a^b$? Question is in title. I (think) I need this to be true to prove a problem for a homework. I'm having trouble proving this with the epsilon definition of limits in a style similar to that of the proofs of the other limit laws. Any help?
 A: Well I don't think this law is at the same level as other limit laws $\lim (a_{n} + b_{n}) = \lim a_{n} + \lim b_{n}$. Hence it would be difficult to go for $\epsilon, \delta$ approach. A better idea is to use the following limit law twice:
If $\lim_{n \to\infty}a_{n} = a$ and function $f(x)$ is continuous at $x = a$ then $\lim_{n \to \infty}f(a_{n}) = f(a)$.
(this law can be proved using $\epsilon, \delta$ type arguments and you should give it a try)
Also in the current question we need to ensure that $a_{n} \to a > 0$. This will ensure that $a_{n} > 0$ after a certain point and this will further ensure that the expression $a_{n}^{b_{n}}$ is defined after a certain point.
Now we know that $a_{n} \to a > 0$ and $\log x$ is continuous at $x = a$ therefore $\log(a_{n}) \to \log a$. And $b_{n} \to b$ so that $b_{n}\log a_{n} \to b\log a$ (product rule for limits).
Again we can see that the function $\exp(x) = e^{x}$ is continuous at $x = b\log a$ and hence by applying the above mentioned rule we get $\exp(b_{n}\log a_{n}) \to \exp(b\log a)$ and this means that $a_{n}^{b_{n}} \to a^{b}$.
