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Math Help - Limit of multivariable function

  1. #1
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    Limit of multivariable function

    I was just wondering if someone could give me short tutorial in limits of multi-variable functions... I understand that you approach the point from an arbitrary line y=cx. but i dont understand when my textbook, and many websites, explain limits using
    0 <  \sqrt{(x-a)^2 + (y-b)^2} < δ

    Thank you in advance
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  2. #2
    MHF Contributor Calculus26's Avatar
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    Recall for the limits of functions of one variable

    |f(x)-L| < e whenever |x-a| < d or a - d < x < a + d

    what you are saying is when the distance from any point x to a < d

    then |f(x)-L| < e

    It's the same basic idea for functions of 2 variables except instead of an interval centered at a you use a circle centered at (a,b)

    so whenever the distance from (a,b) to (x,y) < d then |f(x,y)-L| < e

    using the distance formula this takes the form

    if 0 < < δ then |f(x,y)-L| < e

    Be Careful approaching (a,b) along a line segment y =cx never proves a limit but if you appraoach along 2 different lines, or curves for that matter, and get different results then the limit does not exist.

    hope this helps a little
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  3. #3
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    yes thank you it does... so just one more question
    when you have a function like:
     f(x,y) = ( \frac{x^2y^3}{2x^2+y^2}  if (x,y) DNE (0,0) )
    ( 1 if (x,y) = (0,0) )
    and it asks where the function is continuous.

    Would this just be everywhere except (0,0)???
    Last edited by Tascja; April 16th 2009 at 06:33 PM. Reason: the function didnt make sense
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  4. #4
    MHF Contributor Calculus26's Avatar
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    What is helpful is to realize that sums, products, and quotients of continous functions of one one variable are continuous functions of 2 variables.
    Example x^2 and y^3 are cont fns of one variable so x^2y^3 is a continuous function of two variables etc.

    In your example as long a we stay away from (0,0) we have continuity.

    Suppose you appraoach (0,0) along y = x then f(x,x) = x^5/(3x^2) whose limit is 0. This doesn't mean the limit at(0,0) is 0

    The point is f(0,0) = 1 so even if the limit were 0 f would not be continuos.

    In fact if you approach (0,0) along y= 1/x limf(x,y) doesn't even exist .

    I hope this clarifies things a little.
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