Verification of proof: $f(x) = e^x$ is continuous at $a = 2$

Question

$$|e^x-e^2|<\epsilon$$

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 $\delta=\min{(\delta_1,\delta_2)}$

Answer 1

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

1. 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.
2. 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 $\delta \lt \min(\epsilon / e^3, 1)$.

Answer 2

$e^x$ 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,

$$\delta\le\min(\ln(e^2-\epsilon),\ln(e^2+\epsilon)).$$

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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.