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Limit check

  1. Jun 3, 2005 #1
    I've been doing some revision and I came across a question which boils down to deciding whether or not the following limit is equal to f(0,0) = 0.

    The function is:

    [tex]
    f(x,y) = \left\{ {\begin{array}{*{20}c}
    {x^y ,x > 0} \\
    {0,otherwise} \\
    \end{array}} \right.
    [/tex]

    The limit is:

    [tex]
    \mathop {\lim }\limits_{\left( {x,y} \right) - > \left( {0,y'} \right)} f(x,y) = L
    [/tex]

    where L is either undefined, or defined - I don't know yet and y'(y-prime or y-dash) is a negative number. Approaching (x,y') along any path in the half-plane x <= 0 gives a limit of zero which is fairly obvious. Just looking at the definition of the function I think that the only points which may be discontinuous are (0,y') where y is a negative number.

    I can think of a fudge method to 'show' that the function diverges to infinity if I approach (0,y') from the half plane x > 0(where f(x,y) = x^y) but that obviously isn't sufficient. So I'm hoping that someone can explain to me how to take evaluate the limit(if it exists) of f(x,y) as (x,y) -> (0,y') where y' is a negative number. So for instance how would I evaluate the limit of f(x,y) as (x,y) approaches (0,-5)? I mean, from the half plane x <= 0 the limit is just zero, but what about from the half plane x > 0(where f(x,y) = x^y)? Any assistance would be good, thanks.
     
    Last edited: Jun 3, 2005
  2. jcsd
  3. Jun 3, 2005 #2
    if you're looking for limes of (x,y) to (0,y') then your function is still x^y. make that x^y'. y' is a negative number so you have 1/(x^y). As x -> 0 x^y grows infinitely small, so the whole term 1/(x^y) goes towards infinity. Um... does that help at all?
     
  4. Jun 3, 2005 #3

    arildno

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    Check the behavior of f(x,0) as x goes to 0.
     
  5. Jun 3, 2005 #4
    Thanks for the help so far but could you please give me further assistance?

    f(x,0) is equal to one as I go from very large positive values of x to very small positive values of x. So it seems that I can say f(x,y) has a limit of one if (x,y) -> (0,0). Since this limit is not equal to f(0,0) = 0(by definition of the function) then it is not continuous as (0,0). However, I'm still having trouble dealing with negative y-values. I'm just unsure about how to deal with such values.

    Actually, f(x,y) is a piecewise function so in taking limits along a path(say for instance y = x) which 'part' of the function should I consider? x^y or zero? There are other problems such as, if I take a limit along say y = x, would I only consider y = x for positive x since the function is split into two - half planes x > 0 and x <=0 and f(x,y) is defined differently in those planes.

    Just looking at the graph of x^(b), where b is negative, seems to suggest that as (x,y) -> (0,b) of f(x,y) diverges. I'm not even sure if that is enough to suggest that f(x,y) doesn't converge to a finite value as (x,y) -> (0,b) where b is negative. Perhaps even worse, even if that is true, I can't think of a mathematically 'formal' or 'correct' way of showing that it is indeed the case.

    Edit: Hmm...[tex]x^y = e^{y\log \left( x \right)} = \frac{1}{{e^{ - y\log \left( x \right)} }}[/tex].

    So as (x,y) -> (0,b) , where b is negative f(x,y) approaches something. If I take the limit along the line y = b then x^y -> infinity. I have shown that the limit of f(x,y) -> (0,b) along a single line does not equal f(0,0) = 0 so in this case that is sufficient to say that the function is not continuous at (0,b) where b is negative?
     
    Last edited: Jun 3, 2005
  6. Jun 3, 2005 #5

    arildno

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    Benny:
    Now go along f(0,y)=0 as y goes to 0. The limiting value along this particular path is obviously zero.

    This shows the limit value does not exist at the origin, even if you restrict your attention solely to the function's behaviour in the 1.quadrant.
     
  7. Jun 3, 2005 #6
    Thanks for the help arildno.
     
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