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Why lim x->0 e^-x - 1 / x = -1

  1. Feb 17, 2017 #1
    1. The problem statement, all variables and given/known data
    In the textbook I am studying one of the exercises is: lim x→0 e-x - 1 / x = ? The answer given is -1. But I can't see why this answer should follow.

    2. Relevant equations
    The most relevant equation here is: lim x→0 = ex - 1 / x = 1

    3. The attempt at a solution
    So let's take: lim x→0 e-x - 1 / x = ? So x approximates 0, so you get e-x approximates e-0 which in my view is just e0 = 1. Therefore I don't see how this differs from lim x→0 = ex - 1 / x = 1. In my view the answer should be 1 and not -1... I'm clearly missing something, but I don't have a clue what... Can you please help me?
     
  2. jcsd
  3. Feb 17, 2017 #2

    Mark44

    Staff: Mentor

    You really need parenthese.
    Again, you need parentheses.
    The problem with your analysis is that both the numerator and denominator are approaching 0,
    Your textbook must have examples of problems where both numerator and denominator are approaching 0. Have you looked at them?

    BTW, I agree with the textbook's answer.
     
  4. Feb 17, 2017 #3

    Mark44

    Staff: Mentor

    Thread moved from the Precalc section. Problems about limits are typically at the Calculus level, not the Precalc level.
     
  5. Feb 17, 2017 #4
    Thanks for the tip about the parentheses....

    The problem with the text book I am studying is that it can be a bit lacking in explanation and illustration sometimes.... It's a very comprehensive text book, but also overly concise sometimes...

    Anyway, I appreciate the problem that the denominator is approaching zero... You can't divide through zero... So I probably have to rework the division in such a way that the denominator no longer approximates zero.... I probably have to get back to the chapter on limits, wich I found somewhat hard to follow near the end... I probably missed something there which I need to use here.... But if you could just give me hint in which direction I must look for the answer it would be much appreciated!
     
  6. Feb 17, 2017 #5

    PeroK

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Three suggestions:

    1) Plug in a small value of x into a calculator or spreadsheet and see whether the function is close to ##1## or ##-1##.

    2) What definitions of ##e^{-x}## do you know?

    3) Do you think this might be a "hospital" case?
     
  7. Feb 17, 2017 #6
    Thanks for the tips... I will get to work... Sometimes I really hate that I love mathematics while sucking at it :)
     
  8. Feb 17, 2017 #7

    Mark44

    Staff: Mentor

    Yes, good idea.

    One thing that you seem to be missing is that a limit expression of the form ##[\frac 0 0]## is one of several indeterminate forms. It is indeterminate because the limit itself can be any number, depending on the expression itself.

    Here are three examples where both the numerator and denominator are approaching zero, but with wildly different limiting values.
    1. ##\lim_{x \to 1} \frac {x^2 + 3x - 4}{x - 1}## (The limit is 5.)
    2. ##\lim_{x \to 3} \frac {x^2 - 6x + 9}{x - 3}## (The limit is 0.)
    3. ##\lim_{x \to -2} \frac {x + 2}{x^3 - 6x^2 + 12x + 8}## (The limit is ∞.)
     
  9. Feb 17, 2017 #8
    $$\lim_{x \to 0} \frac {e^x-1}{x }$$ is a special case of $$\lim_{x \to 0} \frac {a^x-1}{x }$$
     
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