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L'Hospital's Rule -X to the power of X

  1. Jan 4, 2013 #1
    1. The problem statement, all variables and given/known data
    [itex]Lim_{x\rightarrow0^{+}}X^{x}[/itex]


    2. Relevant equations



    3. The attempt at a solution

    [itex]ln(x^{x})=xln(x)=\frac{ln(x)}{x^{-1}}[/itex]

    Take the derivative of the top and bottom -

    [itex]\frac{x^{-1}}{-x^{-2}}=\frac{-x^{2}}{x}[/itex]

    So, we're left with: [itex]lim_{x\rightarrow0^{+}}-x[/itex]

    So, the answer would be 0, right? Where di I mess up? My book says the answer should be 1.
     
  2. jcsd
  3. Jan 4, 2013 #2

    jedishrfu

    Staff: Mentor

    Since you used the ln to both sides, I think you forgot the last step to go back to the original equation and plug in the limit answer zero and hence e^0 = 1
     
  4. Jan 4, 2013 #3

    Zondrina

    User Avatar
    Homework Helper

    You forgot to do something. You took the ln, now to balance it out you have to raise something ;)
     
  5. Jan 5, 2013 #4
    Ah, so, we take e to the power of - [itex]lim_{x\rightarrow0^{+}}-x[/itex]

    So you cancel out the ln?
     
  6. Jan 5, 2013 #5

    jedishrfu

    Staff: Mentor

    No, you started with x^x and then converted it to e^xlnx and then you evaluated the limit of xlnx using lhospitals rule

    so that limit was zero hence you have e^0 which is 1 hence x^x as x approaches 0 is 1.
     
  7. Jan 5, 2013 #6
    Understood. Thanks.
     
  8. Jan 5, 2013 #7

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Another way of saying this is that you started with xx but then changed to y= ln(xx)= x ln(x). So what you found was that y[/b[ goes to 0. Because ln(x) is a continuous function that is the same as saying that ln(L)= 0 where "L" is the limit you are actually seeking. And, of course, ln(1)= 0 so L= 1.
     
  9. Jan 5, 2013 #8


    Well, I thought of it like this: [itex]x=e^{ln(x)}[/itex]

    This is ok, right?
     
  10. Jan 5, 2013 #9

    jedishrfu

    Staff: Mentor

    Yes, x = e ^ lnx andtherefore x^x = e ^xlnx
     
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