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Math Help - [SOLVED] Epsilon Delta (limit definition)

  1. #1
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    [SOLVED] Epsilon Delta (limit definition)

    I'm given an epsilon value of 0.5
    The limit of my function evaluated as x-->0 is 1


    my function is

    f(x) = (e^x - 1) / (x)

    Just a restatement of the variable
    e = 0.5
    L = 1
    f(x) = (e^x - 1) / (x)
    x = ?
    d(delta) = ?
    a = 0

    I've plugged the values into the inequalities and i'm stuck on this part:

    |f(x)-L| < e
    i.e.
    | [(e^x - 1) / (x)] - 1| < 0.5
    So, I simplify and then I get to this point:
    e^x - 1 - 1.5x < 0
    I can't figure out x and because I can't figure out x I'm unable to figure out delta. I need some help solving for x. I'm aware that I will likely need to apply a natural logarithm but whenever I do I still am unable to solve for x.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Noxide View Post
    I'm given an epsilon value of 0.5
    The limit of my function evaluated as x-->0 is 1


    my function is

    f(x) = (e^x - 1) / (x)

    Just a restatement of the variable
    e = 0.5
    L = 1
    f(x) = (e^x - 1) / (x)
    x = ?
    d(delta) = ?
    a = 0

    I've plugged the values into the inequalities and i'm stuck on this part:

    |f(x)-L| < e
    i.e.
    | [(e^x - 1) / (x)] - 1| < 0.5
    So, I simplify and then I get to this point:
    e^x - 1 - 1.5x < 0
    I can't figure out x and because I can't figure out x I'm unable to figure out delta. I need some help solving for x. I'm aware that I will likely need to apply a natural logarithm but whenever I do I still am unable to solve for x.
    Note that we want |x - 0| = |x| < \delta. Lets say we want \delta to be at most 1. so |x| < 1 \implies -1 < x < 1 \implies e^{-1} < e^x < e.

    Now, go back to our expression and apply this bound on e^x:

    \bigg| \frac {e^x - 1}x - 1 \bigg| < \bigg| \frac {e - 1}x - 1 \bigg| < 0.5

    no solve the last inequality so that x is bounded between two constants. \delta will be the larger one
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  3. #3
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    Thanks again, I really appreciate your help.
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