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Homework Help: Limit problem

  1. Dec 8, 2018 #1
    1. The problem statement, all variables and given/known data
    ## \lim x-a \frac {{a^x-{x^a}}}{{x^x}-{a^a}} = -1## then a is?

    2. Relevant equations
    L'Hospital rule

    3. The attempt at a solution
    Using LHR we can write numerator as ##\frac{a^x ln{a}-ax^{a-1}}{x^xln(x+1)}##
    plugging x=a and equating to -1 gives 1-ln(a)=ln(a+1); so 1=ln(a(a+1)) but this gives an incorrect answer. Where am I going wrong?
     
  2. jcsd
  3. Dec 8, 2018 #2

    Math_QED

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    Limit of ##x \to a## I assume?
     
  4. Dec 8, 2018 #3
    yes
     
  5. Dec 8, 2018 #4

    Math_QED

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    Maybe this is your mistake:

    ##(x^x)' = x^x(\ln x + 1)##

    Not: ##(x^x)' = x^x \ln(x+1)## like you wrote.
     
  6. Dec 8, 2018 #5
    I routinely make such mistakes while studying math, I know all the correct things and math could be enjoyable and I could feel like I'm good at it if only I stopped making them...do you have any suggestions on how I can stop making these silly mistakes?
    and Thank you very much for your help :D
     
  7. Dec 8, 2018 #6

    Math_QED

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    I'm glad I could help.

    For your question, check out this wonderful article written by someone on this forum who is more experienced/knowledgeable than me, where he adresses the issue (the thing is written for exam situations, but really applies to any situation).

    https://www.physicsforums.com/insights/10-math-tips-save-time-avoid-mistakes/
     
  8. Dec 8, 2018 #7
    That was very helpful, thank you :D
     
  9. Dec 8, 2018 #8

    Ray Vickson

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    You can set ##x = a+h## to write your fraction as
    $$\text{fraction} = \frac{a^{a+h} - (a+h)^a}{(a+h)^{a+h} - a^a}, $$
    then take the limit as ##h \to 0## using l'Hospital's rule. However, my personal preference would be to expand the numerator and denominator in a few powers of small ##|h|##. For example, the numerator is ##N_h = a^a a^h - (a+h)^a.## We can expand the first term by setting ##a = e^{\ln a}##, so that ##a^h = e^{h \ln(a)}##, which is easily expanded in powers of ##h##; keeping only linear terms in ##h## is good enough in this problem. The second term is ##(a+h)^a,## which has a binomial expansion in ##h##. Again, stopping at terms linear in ##h## is good enough in this problem. The denominator involves ##(a+h)^{a+h},## which can be written as ##e^{(a+h) \ln(a+h)} ##. You can expand the exponent ##r = (a+h) \ln(a+h)## in powers of ##h## (again, terms linear in ##h## are good enough); then you can expand the exponential ##e^r## as ##1 + r + \cdots##, where ##\cdots## stands for higher powers in ##r##. Now substitute your expansion of ##r## in terms of ##h##.

    Of course, in principle, this "expansion" method is really just l'Hospital's rule in disguise, but I find it faster and easier than direct application of l'Hospital. (However, it all depends on how familiar you are with the expansions of the standard functions.)
     
  10. Dec 9, 2018 #9
    I have practiced mostly with L'Hospital, although I'm aware of the expansions of standard functions. (The trouble is I can't remember and retain them...)
     
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