# 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?

• I think you need some assumption on $a,b$ for this to make sense, say $0 \leq a,b < \infty$, so that you don't run into complex numbers or division by zero. What happens when $a=0$ or $b=0$? – Jair Taylor Feb 24 '14 at 1:37
• Can we assume that $a,b \in \mathbb{R}$ and that one of $a$ and $b$ is non-zero? – Omnomnomnom Feb 24 '14 at 1:37
• This is just continuity of the exponentiation operation. – Berci Feb 24 '14 at 1:38
• – mle Feb 24 '14 at 1:45
• If you've been told to use the $\epsilon$-$\delta$ definition, then so be it. But I would prove that $\lim_{n\to\infty} b_n\log a_n = b\log a$ and then exponentiate. This uses the continuity of exponentiation (and of the logarithm, to get $\lim_{n\to\infty} \log a_n =a$ in the first place), as well as the limit of the product of two sequences which presumably is available to you. – Greg Martin Feb 24 '14 at 2:49

## 1 Answer

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 $a_{n}^{b_{n}} \to a^{b}$.