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Math Help - Existance of limit

  1. #1
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    Existence of limit

    Hey!
    Imagine a function f of multiple variables. Does there have to be some neighborhood around a point p in which f is defined for all points for  \lim_{x \rightarrow p} f(x) to exist? As a concrete example, does the following limit exist?

    \lim_{(x,y) \rightarrow (0, 0)} sin(x)/x + y
    Last edited by thesmurfmaster; June 12th 2012 at 10:01 AM.
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  2. #2
    mfb
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    Re: Existance of limit

    The limit exists, and it is independent of the definition of the function at (0,0) (which has to be different from sin(x)/x+y, if the function is defined there).

    Note that your function definition is incomplete, you did not provide the domain and codomain of the function.
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  3. #3
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    Re: Existance of limit

    OP, do you mean \lim_{(x,y) \rightarrow (0, 0)} \sin(x)/x + y or \lim_{(x,y) \rightarrow (0, 0)} sin(x)/(x + y)?

    Quote Originally Posted by mfb View Post
    The limit exists, and it is independent of the definition of the function at (0,0)
    The question is not about (0, 0) (we all know that the function value at the limit point is irrelevant to the definition of the limit), but about points (x, y) where x = -y (if the function is indeed sin(x) / (x + y)).

    Quote Originally Posted by mfb View Post
    Note that your function definition is incomplete, you did not provide the domain and codomain of the function.
    It's reasonable to assume that the domain consists of all points where the ratio sin(x) / x (or sin(x) / (x + y)) is defined.

    Quote Originally Posted by thesmurfmaster
    Does there have to be some neighborhood around a point p in which f is defined for all points for \lim_{x \rightarrow p} f(x) to exist?
    According to Wikipedia and Encyclopedia of Mathematics, the answer is no. The definition of limit takes the intersection of a neighborhood of the limit point and the domain of the function.

    That said, \lim_{(x,y) \rightarrow (0, 0)} sin(x)/(x + y) does not exist. Indeed, the limit is 1 when (x, y) approaches (0, 0) along the x-axis. On the other hand, every open circle around (0, 0) has points (x, y) where |sin(x) / (x + y)| is arbitrarily large (fix some x and make y approach -x).
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    Re: Existance of limit

    Quote Originally Posted by emakarov View Post
    OP, do you mean \lim_{(x,y) \rightarrow (0, 0)} \sin(x)/x + y or \lim_{(x,y) \rightarrow (0, 0)} sin(x)/(x + y)?

    The question is not about (0, 0) (we all know that the function value at the limit point is irrelevant to the definition of the limit), but about points (x, y) where x = -y (if the function is indeed sin(x) / (x + y)).

    It's reasonable to assume that the domain consists of all points where the ratio sin(x) / x (or sin(x) / (x + y)) is defined.

    According to Wikipedia and Encyclopedia of Mathematics, the answer is no. The definition of limit takes the intersection of a neighborhood of the limit point and the domain of the function.

    That said, \lim_{(x,y) \rightarrow (0, 0)} sin(x)/(x + y) does not exist. Indeed, the limit is 1 when (x, y) approaches (0, 0) along the x-axis. On the other hand, every open circle around (0, 0) has points (x, y) where |sin(x) / (x + y)| is arbitrarily large (fix some x and make y approach -x).
    Not to be crass, but it should be quite obvious that I meant what I wrote, since sin(x)/(x+y) doesn't illustrate my point at all as the limit of it wouldn't exist either way! Thanks for the article in Encyclopedia of Maths though, it set my mind at rest.
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    Re: Existance of limit

    Since you asked whether the function has to be defined in a whole neighborhood of p (and not just at p) I thought that the problematic situation was when every neighborhoods has points not in the domain of the function. I think whether the definition of limit allows such situations is a more interesting question. In fact, before I looked up the links above, I thought that the function has to be defined in some punctured neighborhood of p. This is why I assumed the denominator was (x + y).

    It is well known that the function does not have to be defined at p in order to take its limit at p. This is equally true for single-variable and multi-variable functions.
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    Re: Existance of limit

    Quote Originally Posted by emakarov View Post
    Since you asked whether the function has to be defined in a whole neighborhood of p (and not just at p) I thought that the problematic situation was when every neighborhoods has points not in the domain of the function. I think whether the definition of limit allows such situations is a more interesting question. In fact, before I looked up the links above, I thought that the function has to be defined in some punctured neighborhood of p. This is why I assumed the denominator was (x + y).
    I'm not sure we're on the same page. The function I wrote leaves the question just the same as it was, with the added bonus that the limit does indeed exist if one can disregard that it isn't defined in every point of any neighborhood of (0, 0). The only difference is that my function isn't defined when x=0, and yours is not when x=-y . Adding y is quite unnecessary really, other than to highlight it as a function of two variables.

    Quote Originally Posted by emakarov View Post
    It is well known that the function does not have to be defined at p in order to take its limit at p. This is equally true for single-variable and multi-variable functions.
    Obviously!
    Last edited by thesmurfmaster; June 12th 2012 at 10:03 AM.
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