Proving that $e^x$ is continuous.

Question

I just have one quick, and perhaps slightly trivial question.

Why does it suffice to prove that $e^x$ is continuous at $x=0$ to prove that it is continuous over all of the reals?

Answer 1

$e^{k}=e^{k+0}=e^k\cdot e^0=\text{some constant}\times e^0$. So if it can be proved to be continuous at $0$ , it will be continuous for all reals.

WHY ? . If you know how to show that $e^x$ is continuous at $0$, you will know the reason!

Answer 2

If $e^x$ is continuous at $x=0$ then $$\forall \epsilon>0 \exists\delta>0: |x|<\delta\implies |e^x-1|<\epsilon.$$

So $$\forall \epsilon>0 \exists\delta>0: |x-x_0|<\delta\implies |e^x-e^{x_0}|=e^{x_0}|e^x-1|<e^{x_0}\epsilon,$$ which shows continuity at $x_0.$