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**james121515** Hi guys,

I have a question about an $\displaystyle \epsilon / \delta$ proof, specifically, properly handling the inequality at the end.

Prove that $\displaystyle \lim_{x \to 2} e^x = e^2$

Let $\displaystyle \epsilon > 0$ be given. Let $\displaystyle \delta = \ln(\frac{\epsilon}{e^2}+1)$.

(1) Then $\displaystyle |x-2| < \ln(\frac{\epsilon}{e^2}+1)$ implies

(2) $\displaystyle e^{|x-2|} < e^{\ln{\frac{\epsilon}{e^2}+1)}$ implies

(3) $\displaystyle e^{|x-2|} < \frac{\epsilon}{e^2} + 1$ implies

(4) $\displaystyle e^{|x|} =e^x < \epsilon + e^2$

We know that $\displaystyle e^x > 0, e^2 > 0$ for all $\displaystyle x \in \mathbb{R}$, hence (4) above implies

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

Now, my question involves this line. Clearly this line above is not equivalent to $\displaystyle |e^x - e^2| < \epsilon $, which is what we're trying to show. For my formal proof to be complete, how do I go from here?

Could I say something like:t

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

Is that even a true statement, and if so, is that how you do this proof?

Thanks for your help.