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Limits involving e and ln

  1. Mar 3, 2012 #1
    1. The problem statement, all variables and given/known data
    Show that lim (x,y) → (a,0) e^(x ln y) = 0 [itex]\forall[/itex]a > 0


    2. Relevant equations



    3. The attempt at a solution
    I've tried looking at lim (x,y) → (a,0) x ln y seperately.
    lim(x,y) → (a,0) x ln y = lim(x,y) -> (a,0) x * lim(x,y) → (a,0) ln y
    = a * lim(x,y) → (a,0) ln y

    Now lim(x,y) → (a,0) ln y is -∞. So we get lim(x,y) →(a,0) x ln y = -∞. Now lim z → -∞ e^z = 0.

    The problem here is that we haven't discussed limits involving infinity in class and i'm pretty sure i'm not allowed to use it. My question therefore is: is there any other way to show that lim (x,y) → (a,0) e^(x ln y) = 0?
     
  2. jcsd
  3. Mar 3, 2012 #2

    Dick

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    No, I think that's the right way to do it.
     
  4. Mar 3, 2012 #3
    That's what I was afraid off :(. How would I prove that lim (x,y) → (a,0) e^(x ln y) = 0 and lim z → -∞ e^z = 0? Im familar with the [itex]\delta[/itex] and [itex]\epsilon[/itex] method but im not sure how to include ∞ in this method.
     
  5. Mar 3, 2012 #4

    Dick

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    lim z → -∞ e^z = 0 just means that for any ε>0 you can find a N such that e^z<ε if z<N. Try picking N=log(ε).
     
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