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Homework Help: L'Hospital's Rule

  1. Dec 12, 2009 #1
    Find the following limit using L'Hospital rule..

    lim..........(e^x) / [(e^x) + 7x]
    x->infinity

    = f'(x) / g'(x)
    =e^x / (e^x) + 7
    = infinity/infinity
    =0
    am i got it right?
     
  2. jcsd
  3. Dec 12, 2009 #2

    HallsofIvy

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    Science Advisor

    Okay to here

    What does that mean?

    Why would you think that?

    One you have e^x/(e^x+ 7) either use L'Hospital's rule again or divide both numerator and denominator by e^x.
     
  4. Dec 12, 2009 #3
    e^x/e^x = 1 ???
     
  5. Dec 12, 2009 #4

    CompuChip

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    Homework Helper

    Err... wait. You said: when I take the limit directly I get infinity / infinity, which is not defined. So we need L'Hôpitals rule. I agree with that.
    Then you take the derivatives, and get infinity / infinity again. Then suddenly, that is defined and equal to 0?

    The idea is, that
    [tex]\lim_{x \to \infty} \frac{f(x)}{g(x)} = \lim_{x \to \infty} \frac{f'(x)}{g'(x)} [/tex]
    in this case, but on the right hand side you get something which is still of the form "infinity / infinity". So you will need to apply it again.

    Afterwards, to check your answer, you could try multiplying numerator and denominator by exp(-x).
     
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