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Determine the limit of this function

  1. Apr 12, 2013 #1
    1. The problem statement, all variables and given/known data
    find the limit of [itex]x^x[/itex]when x approaches 0.


    2. Relevant equations



    3. The attempt at a solution
    I am currently revising my Mathematical Analysis and this question pops up. The author only expects the reader to use the most fundamental knowledge of the limit, so I feel my method counts as "cheating"(rewriting as [itex]e^{x lnx}, lim x lnx= 0[/itex]using the L'Hospital's Rule so the limit is 1). Is there an approach that involves "less complicated knowledge" because I really wish to fully grasp every notion of basic math? Thank you...
     
  2. jcsd
  3. Apr 12, 2013 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    That's perfectly valid- you are using the the fact that exponential is a continuous function. That is, [itex]\lim_{x\to 0} e^{f(x)}= e^{\lim_{x\to 0} f(x)}[/itex]. [itex]x^x= e^{x ln(x)}[/itex] and [itex]x ln(x)[/itex] goes to 0 as you say. The limit of [itex]x^cx[/itex] is [itex]e^0= 1[/itex].
     
  4. Apr 12, 2013 #3
    Thanks for the reply. That's my idea exactly. But that problem appears in a chapter well before the author introduces the L'Hospital's Theorem so there must be another way to do it... I am torturing myself...
     
  5. Apr 13, 2013 #4

    Mark44

    Staff: Mentor

    I would check to see if the author has discussed this limit in the chapter you're in: ## \lim_{x \to 0} x ln(x)##. Since he doesn't present L'Hopital's Rule until later, it's possible he has said something about this limit, which is the crux of the problem you're trying to solve.
     
  6. Apr 13, 2013 #5
    You don't really need l'Hopital for [itex]
    \lim_{x\rightarrow 0} {xlnx}
    [/itex]. I would let [itex] x = e^{-n} [/itex] and take the limit as n→∞.
    [itex] \lim_{n\rightarrow \infty} {e^{-n}ln(e^{-n})} = \lim_{n\rightarrow \infty} {\frac {-n}{e^n}} [/itex]. I guess this limit is technically a l'Hopital problem, but it isn't hard to see that it goes to zero.
     
  7. Apr 16, 2013 #6
    Thank you, HS-Scientist, that's what I later did. It is a bit tedious because jumping from n to x also takes an extra proof so maybe in the end L'Hopital law is easier.
     
  8. Apr 16, 2013 #7

    Mark44

    Staff: Mentor

    It's not obvious to me that the latter limit is zero. This limit is an indeterminate form [-∞/∞], which means that L'Hopital's Rule applies, so how did you evaluate the limit without using L'H?
     
  9. Apr 16, 2013 #8
    If your definition of e^x is its power series, then this is rather trivial. You can also use the idea that e^x is increasing faster than x without referencing l'Hopital. If we consider [itex] \frac{x}{e^x} [/itex] and replace x with x+r for any positive r>ln2 we get [itex] \frac{x+r}{e^re^x} [/itex]. The numerator has increased by a factor of 1+r/x and the denominator by a factor of e^r, so we multiplied the fraction by [itex] \frac{1+r/x}{e^r} [/itex]. Now for all x>(1+r)/2 (we can assume that this is the case because r is a constant and we can make x as large as we like), this ratio is smaller than 2/e^r. As long as we have r>ln2, this ratio is also less than one. So each time we increase x by r, we multiply it by something that is smaller than a constant which is less than one, so it must go to zero as we add r to the input an arbitrarily large number of times.

    Edit: This argument makes an assumption which I forgot to state explicitly; namely that if the sequence [itex] a_1,a_2,...a_n... [/itex] goes to zero, then so does a function f such that [itex] f(n)=a_n [/itex] (In my argument, a_n is n/e^n evaluated at x,x+r...) This is not true in general, but here it works because of r being arbitrary (as long as it >ln2). This is because f(x) is always bounded from above by the smallest value of f more than ln2 less than x. So every time x is incremented by r, there is a new maximum value of f (smaller than the previous one) for all x greater than this new x.
     
    Last edited: Apr 16, 2013
  10. Apr 17, 2013 #9
    My method is similar to HS-Scientist. It first starts from the statement that [itex]lim\frac{n}{a^n}=0[/itex] where a > 0 and n is an integer, whose proof uses the fact that the sequence is bounded below and [itex]lim \frac{a^{n+1}}{a^n}=\frac{1}{a}<1[/itex]. And then it passes to the continuous case so for any x we can take the integer part of it and construct two terms which sandwich x/e^x and have a common limit 0. Hence x/e^x approaches 0. But indeed it is too tedious..
     
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