How do I prove that $\lim _{x\to a}e^x = e^a$?

Question

I want to be able to do this without assuming that the derivative of $e^x$ is $e^x$. I've tried to do this using both the delta-epsilon definition and the squeeze theorem, but I'm still stuck.

Answer 1

Since $a \in \mathbb{R}$ is constant, note that $$ \lim_{x \to a} e^x = e^a $$ is equivalent to $$ \lim_{x \to a} e^{x-a} = 1 $$ which is equivalent to $x-a \to 0$ since $e^x=1$ has only one solution...

Answer 2

It is true by continuity, indeed note that by esponential properties

$$ \lim_{x \to a} e^x = \lim_{h\to 0} e^{h+a}=e^a\cdot \lim_{h\to 0} e^{h}=e^a\cdot1=e^a$$