Prove ${\lim_{x \to a} e^{x} = e^{a}}$ with a $\delta-\epsilon$ argument

Question

$\displaystyle{|x-a|< \delta}$

$\displaystyle{|e^{x} - e^{a}|< \epsilon}$
$⟺\displaystyle{\ln|e^{x} - e^{a}|< \ln\epsilon}$ $⟺\displaystyle{\frac{x}{a} < \ln\epsilon}$

if $a$ > 0

then $\displaystyle{x < a\ln\epsilon}$ $⟺\displaystyle{x-a<a (\ln\epsilon - 1)}$ $⟺\displaystyle{x-a \leq |x-a| < \delta}$

Choose $\delta$ < $a\ln\epsilon$ such that $x-a<a (\ln\epsilon - 1)$

if $a$ < 0

then $\displaystyle{x > a\ln\epsilon}$ $⟺ \displaystyle{x-a>a (\ln\epsilon - 1)}$ $⟺ \displaystyle{\delta > x-a \geq |x-a| }$

Choose $\delta$ < $a\ln\epsilon$ such that $x-a>a (\ln\epsilon - 1)$

Is this correct?