Attempt on proving $e^x$ is continuous on $\Bbb{R}$

Question

I have seen some proofs of the continuity of $e^x$ (on $\Bbb{R}$), and I would like to see if my take on this proof is correct.
For any $\epsilon>0$, we choose $\delta=\ln(\frac{\epsilon}{e^c}+1)$, and $x,c\in\Bbb{R}$. For $0<|x-c|<\delta$, we have
\begin{align} |e^x-e^c| & = e^c|e^{x-c}-1|\\ & < e^c|e^{\ln(\frac{\epsilon}{e^c}+1)}-1|\\ & = e^c|\frac{\epsilon}{e^c}+1-1|\\ & = e^c|\frac{\epsilon}{e^c}|\\ & = \epsilon \end{align} $\therefore |e^x-e^c|<\epsilon$ when $0<|x-c|<\delta$ and $\delta=\ln(\frac{\epsilon}{e^c}+1)$, which implies that $e^x$ is continuous on $\Bbb{R}$.
Any corrections on this proof is appreciated.

Answer 1

This works if $x\gt c$. If $x\lt c$, consider $|e^c-e^x|$ instead.