Elementary proof of $\lim\limits_{x\to 0}a^x = 1$ when $a>0$

Question

I have a proof which uses sequences that converge to $0$. I think there is an easier proof than mine but I couldn't find. Is there anyone can prove this without using sequences? Or would you please give me some links? Epsilon-delta argument needed.

Answer 1

Fix $a \neq 0$, then $f(x) = a^x$ is a continuous function. Then $\lim_{x \rightarrow 0} f(x) = f(0) = a^0 = 1$. You have to prove two things:

1. the limit of the sequence is the limit of the corresponding continuous functions (in $\mathbb{R}$);
2. $a^x$ is continuous.

To prove (1):

• the limit for sequences is $$\forall \epsilon>0, \ \exists \tilde{n} \in \mathbb{N}, \ \forall n>\tilde{n}, \ |a_n - a_{\tilde{n}}| < \epsilon$$

• the limit for continuous functions (in $\mathbb{R}$) is $$\forall \epsilon > 0, \ \exists \delta > 0, |x-y| < \delta \Rightarrow |f(x)-f(y)| < \epsilon$$

If you already know that $f(x)$ is continuous in $0$, then calling $a_n = f(\frac{1}{n})$ you have $\lim_{x \rightarrow 0} f(x) = \lim_{n \rightarrow \infty} a_n$, where the first is a limit of continuous functions, the second is of sequences.

To prove (2): you have to prove that the exponential is continuous. You can find some proofs in this site, like here.