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Math Help - limit of exponential function

  1. #1
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    limit of exponential function

    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.
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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 \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
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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:

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