limit of exponential function

**james121515** · November 6th 2011, 07:34 PM

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

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

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

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

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

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

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

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

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

Now, my question involves this line. Clearly this line above is not equivalent to $|e^x - e^2| < \epsilon$ , 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

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

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

Thanks for your help.

**Prove It** · November 6th 2011, 08:06 PM

Originally Posted by james121515

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

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

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

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

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

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

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

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

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

Now, my question involves this line. Clearly this line above is not equivalent to $|e^x - e^2| < \epsilon$ , 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

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

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

Thanks for your help.

You're on the right track, but you need to use the reverse triangle inequality.

$\displaystyle |a| - |b| \leq |a - b|$ .

So what is happening in your inequality is

$\displaystyle \begin{align*} |x| - |2| &\leq |x - 2| \\ |x| - 2 &\leq |x - 2| \\ e^{|x| - 2} &\leq e^{|x - 2|} \\ e^{|x|}e^{-2} &\leq e^{|x - 2|}\end{align*}$

and since we already knew $\displaystyle e^{|x - 2|} < \frac{\epsilon}{e^2} + 1$ , that means

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

and you should be able to go from there

Edit: OK this will technically only prove the right-hand limit, where $\displaystyle |x| > 2$ . To prove the left-hand limit, you will need to reverse the terms on the LHS...

$\displaystyle |2| - |x| \leq |x - 2|$

and you should get the same solution

**Deveno** · November 6th 2011, 08:24 PM

i think we only need to do this for the left-hand limit. for the right-hand limit, from (4) we have:

$|e^x - e^2| = e^x - e^2 < \epsilon$ , since $e^x$ is increasing.

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