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Simple limits question.

  1. Oct 20, 2007 #1
    I just have a question of principles.
    I know [tex]0^0[/tex] and [tex]\infty^0[/tex] are both undefined.
    But isn't this true:
    [tex]\lim_{x \to 0}x^0 = 1[/tex] and [tex]\lim_{x \to \infty}x^0 = 1[/tex]?
     
  2. jcsd
  3. Oct 20, 2007 #2

    arildno

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    Certainly.
    Now, compute:
    [tex]\lim_{x\to{0^{+}}}0^{x}[/tex] and [tex]\lim_{x\to\infty}(2^{x})^{\frac{1}{x}}[/tex]
     
    Last edited: Oct 20, 2007
  4. Oct 20, 2007 #3
    I just argued the same thing in of your posts. I thought it should be. But i guess thats only if we take the x to be neglegible..

    Edit: nvm just ignore me..
     
    Last edited: Oct 20, 2007
  5. Oct 20, 2007 #4
    Thanks, this question did arise from this thread: https://www.physicsforums.com/showthread.php?t=192598

    Since [tex]x^0 = \frac{x}{x}[/tex], I would go as far as saying [tex]\infty^0=\frac{\infty}{\infty}=1[/tex] without the limits... as long as the infinities are the same, and they should be since that is what the equation requires. But again, messing with infinities is not my thing ;)
     
  6. Oct 20, 2007 #5

    arildno

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    Read again my comment:
     
  7. Oct 20, 2007 #6

    For the first one, [tex]\lim_{x\to{0^{+}}}0^{x} = 0[/tex]. However, I can do this only intuitively, by mentally graphing the equation. Is there an algebraic way?

    And for the second one,
    [tex]\lim_{x\to\infty}(2^{x})^{\frac{1}{x}} = \lim_{x\to\infty}2^{x^0} = 2^1 = 2[/tex].
     
  8. Oct 20, 2007 #7

    arildno

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    Indeed correct!

    But, wouldn't you agree to that the first limit is of the shape [tex]0^{0}[/tex], and the other of the shape [tex]\infty^{0}[/tex]?
     
  9. Oct 20, 2007 #8
    Thank you very much arildno, your point is well-made!
     
  10. Oct 20, 2007 #9

    arildno

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    So, as you can see, for symbol groups like [tex]0^{0},\infty^{0}[/tex], and for that matter, [tex]1^{\infty},\frac{\infty}{\infty},0*\infty,\infty-\infty,\frac{0}{0}[/tex], we cannot relate to such a group a unique number so that all limiting processes "tending" towards that symbol group will converge to that number.


    That is why we prefer to say these expressions are "indeterminate", or "undefined".
     
  11. Oct 20, 2007 #10
    Try this one!

    [tex]\lim_{t\to 0} (\cos (x))^{-\frac{1}{x^2}}[/tex]

    The answer may surprise you.

    And for the daring:

    [tex]\lim_{t\to 0} (\cos (x))^{-\frac{1}{x^n}}[/tex]
     
    Last edited: Oct 20, 2007
  12. Oct 20, 2007 #11
    Found this forum yesterday. Just registered and here's my first post.
    The solution to the first one is $\sqrt{e}$. Write the function as an exponential
    function and take the limit inside. Using 'l Hopital gives 1/2.

    The second one uses the same principal, however the solution is depending on
    the value of n. Assuming it can only take on integer numbers (pos and neg) we have
    for the negative ones always 1. For n=0 and n=1 it is also 1. The equation with n=2
    is allready given. For values of n higher than 2 the limit is infinity.

    Hopefully the latex thing works.
     
  13. Oct 20, 2007 #12

    arildno

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    [tex]\lim_{t\to 0} (\cos (x))^{-\frac{1}{x^2}}=(\cos (x))^{-\frac{1}{x^2}},x\neq{0}[/tex]

    Not really.
    The same procedure..
     
  14. Oct 20, 2007 #13
    You're assuming that constant functions are continuous.
     
  15. Oct 20, 2007 #14
    It seems there is something not right.
    *) The limit for t? It isn't in the function.
    *) How do you display the formulas the latex way?
     
  16. Oct 20, 2007 #15

    arildno

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    No, I'm assuming that t is a different and independent variable from the x variable.
     
  17. Oct 20, 2007 #16
    Independence implies constancy, as otherwise x would be a function of t. That is, f(x) is constant in t if x and t are independent.

    But whatever. You know what I meant and my example was another illustration of your point.
     
    Last edited: Oct 20, 2007
  18. Oct 20, 2007 #17

    arildno

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    No, what you said was that I specifically assumed that constant functions are CONTINUOUS.

    That is:

    a) A trivial fact, and not an assumption.

    b) Irrelevant to my post, or indeed, to your example.



    As limit exercises taking x to zero, rather than t, your examples are nice, though.
     
    Last edited: Oct 20, 2007
  19. Oct 23, 2007 #18
    Hmm... do you mean [tex]\lim_{x\to0}[/tex]?
     
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