Verification of proof: $f(x) = e^x$ is continuous at $a = 2$ $$|e^x-e^2|<\epsilon$$
So if $x<2$ then $2-x<\delta $
Calculations:
$$e^2-e^x<\epsilon$$
$$-e^x<\epsilon -e^2 $$
$$e^x>e^2-\epsilon$$
$$x>\ln(e^2-\epsilon)$$
$$-x<-\ln(e^2-\epsilon)$$
$$2-x<2-\ln(e^2-\epsilon)$$
Then we get:
$$2-x<2-\ln(e²-\epsilon) = \delta_1 $$
And then for $x>2$:     
$x-2<\delta$
$$e^x-e^2<\epsilon $$
Calculations:
$$e^x<\epsilon+e^2 $$
$$x<\ln(\epsilon+e^2)$$
$$x-2<\ln(\epsilon+e^2)-2$$
Then we get:
$$x-2<\ln(\epsilon+e^2)-2=\delta_2$$
So that means $\delta=\min{(\delta_1,\delta_2)}$
 A: $e^x$ is a monotonic function. So if we solve
$$e^{2\pm\delta}=e^2\pm\epsilon,$$ (assuming $e^2>\epsilon$) the smallest solution in $\delta$ will indeed do.
Hence,
$$\delta\le\min(\ln(e^2-\epsilon),\ln(e^2+\epsilon)).$$

Anyway, these bounds look complicated as they involve logarithms. We can simplify using the inequality $\dfrac{x-1}x<\ln x<x-1$:
$$\ln(e^2+\epsilon)=2+\ln(1+\frac\epsilon{e^2})>2+\frac\epsilon{e^2+\epsilon}.$$
$$\ln(e^2-\epsilon)=2+\ln(1-\frac\epsilon{e^2})<2-\frac\epsilon{e^2}.$$
Finally, a suitable elementary bound is
$$\delta\le\min(\frac\epsilon{e^2+\epsilon},\frac\epsilon{e^2})=\frac\epsilon{e^2+\epsilon}.$$
Below, the exact bounds in black & green, and approximate, tighter bounds in khaki & yellowish.

A: Overall, your proof is working. There is however some important pitfalls:


*

*You write a succession of inequalities without stating if those are equivalent. This is however critical if you want to go reverse, i.e. if $\vert x-2 \vert < \delta$ then $\vert e^x - e^2 \vert < \epsilon$ which is what is at the end required.

*You can't take the logarithm of negative numbers. This is the case for example when $\epsilon > e^2$. You have to deal with those cases.


As you know the Mean value theorem, a prove can be:
It exists $c \in (x, 2)$ such that:
$$\vert e^x - e^2\vert = e^c \vert x - 2 \vert \le e^3 \vert x - 2 \vert$$ providing that $\vert x - 2 \vert \lt 1$. And also
$$\vert e^x - e^2\vert = e^c \vert x - 2 \vert \le e^3 \vert x - 2 \vert \le \epsilon$$ if $\vert x - 2 \vert \lt \delta$ for $\delta \lt \min(\epsilon / e^3, 1)$.
