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Homework Help: Quick limit question

  1. Feb 20, 2010 #1
    limit as x approaches 0 from the right of [e^x-(1+e^lnx)]/x^3

    so I tried to plug in .0001 to just see what it the function is doing and I got:
    0/(1*10^-12)
    so I assumed that it approach 0. I just want to make sure that my logic is ok.
     
  2. jcsd
  3. Feb 20, 2010 #2
    actually, I see it a different way. e^lnx is just x, so the function is now lim as x --> 0 of
    [e^x -1 -x]/x^3. See the indeterminate form 0/0, so use l'hopital's rule.
    now the function becomes [e^x - 1]/3x^2, 0/0 indeterminate form, so apply l'hopital's rule again. [e^x]/6x. If you plug in zero, you'll get 1/0, but since you're taking the limit, you can say the function approaches infinity when x approaches zero.
     
  4. Feb 20, 2010 #3
    I thought of doing that originally, but then I looked at the graph, and it does appear that when you take the limit as x approaches 0 from the right, it is getting closer and closer to 1.
     
  5. Feb 20, 2010 #4
    Well, the answer using L'hopital's rule isn't always the same as the answer from taking the limit without L'hopital's rule, so you are more likely to be right if what you said is true.
     
  6. Feb 20, 2010 #5
    so...don't l'hopital it?
     
  7. Feb 20, 2010 #6

    Mark44

    Staff: Mentor

    I think you might have made a mistake entering something when you got the graph. As x approaches 0 from the right, the expression does not approach 1.
     
  8. Feb 21, 2010 #7
    Yeah, I made a mistake, it doesn't approach infinity from my logic. Sorry about that.
     
  9. Feb 21, 2010 #8

    Mark44

    Staff: Mentor

    That's not true. However you take the limit you should get the same result. If you get a different result using L'Hopital's Rule, it's very likely that you are applying it to a limit to which it doesn't apply.
     
  10. Feb 21, 2010 #9
    Hold on, correction to what I said earlier, I got the graph of this problem and other problem messed up. But either way, I don't understand if I should talk the limit as is (which gets me 0, which can't be right looking at the correct graph) or L'hopital the problem.
     
  11. Feb 21, 2010 #10

    Mark44

    Staff: Mentor

    If you use L'Hopital's Rule twice, you finally get to a limit that L'Hopital's Rule cannot be used on. In that case, you get an expression whose numerator approaches 1, and whose denominator approaches 0 (from the positive side, since x is approaching zero from the positive side). This limit agrees completely with the behavior shown by the graph.
     
  12. Feb 21, 2010 #11
    ah. SO I then get 1/0 or that it goes to inf. While I have the answer now, can you explain how you know that you need to use l'hopital, because all of the examples that I have seen have not involved a left of right handed limit.
     
  13. Feb 21, 2010 #12

    Mark44

    Staff: Mentor

    You can use L'Hopital's Rule if your limit is of the indeterminate form [0/0] or [+/- infinity/infinity]. There are several other indeterminate forms, such as [1^infinity] and [infinity - infinity]. L'Hopital's Rule doesn't apply to those. It also does not apply to expressions where the numerator approaches a nonzero constant, but the denominator approaches zero.

    Clear?
     
  14. Feb 21, 2010 #13
    Yes, and even if you have a left or right sided limit you can still use L'Hopital's rule?
     
  15. Feb 21, 2010 #14

    Mark44

    Staff: Mentor

    Sure, that makes no difference. The main reason for the right-sided limit in this problem is that eln x is defined only for x > 0.
     
  16. Feb 21, 2010 #15
    Oh, Ok that makes sense now.
     
  17. Feb 21, 2010 #16
    That's what I meant to say. I wanted to say that sometimes L'Hopital's Rule doesn't apply, but I said it the wrong way. Thanks.
     
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