1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Limits probably involving e^x-1/x special limit

  1. Sep 29, 2013 #1
    1. The problem statement, all variables and given/known data
    lim of (a^x-a^-x-2)/(x^2) as x->0

    Find the value of the limit^.


    The answer is (lna)^2
    I know how to get the answer using L'Hospital.
    However, I do not want to use Hospital's rule.
    I think I see the pattern with the special limit:
    lim of (e^x-1)/x as x->0

    But I just don't see how I can apply that. I tried to take ln() out of everything, but it leads me to nowhere.
    Please help me.
     
  2. jcsd
  3. Sep 29, 2013 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    a^x=e^(ln(a)x) is the connection between powers of e and a. You could also substitute the series expansions of the function. But I don't think knowing the special limit you've got there will help.
     
  4. Sep 29, 2013 #3
    Then, how else could we solve this problem?
    My school hasn't taught me Hospital's rule. So, there must be another way.
     
  5. Sep 29, 2013 #4
    As far as I can see, L'Hopital's rule doesn't even apply since the numerator does not equal 0 if you plug in 0. a^0 - a^(-0) - 2 = 1 - 1 - 2 = -2.

    You're simply dividing by a small positive number, so the limit is negative infinity.
     
  6. Sep 29, 2013 #5

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    I think there's a typo in the original post. The problem should be
    $$\lim_{x \to 0} \frac{a^x + a^{-x} - 2}{x^2}.$$ You can rewrite this as
    $$\lim_{x \to 0} \frac{a^{-x}[(a^x)^2 - 2a^x +1]}{x^2}.$$ You can do a little algebra, use what Dick said about expressing this in terms of ##e##, and change variables appropriately to relate it back to the special limit.
     
  7. Sep 29, 2013 #6

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Ok, that's nice. Now I see how to use the 'special limit'.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Limits probably involving e^x-1/x special limit
  1. Limit of (E^x+1)^1/x (Replies: 3)

  2. 1/(e^x-1) limit (Replies: 6)

  3. Limit of arctan(e^x)? (Replies: 3)

Loading...