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Thread: limit of exponential function

  1. #1
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    limit of exponential function

    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.
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  2. #2
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    Re: limit of exponential function

    Quote Originally Posted by james121515 View Post
    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.
    You're on the right track, but you need to use the reverse triangle inequality.

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

    So what is happening in your inequality is

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

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

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

    and you should be able to go from there

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

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

    and you should get the same solution
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  3. #3
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    Re: limit of exponential function

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

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