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Thread: Functions of Several Variables

  1. #1
    Apr 2008

    Functions of Several Variables

    Let F: $\displaystyle R^2 \rightarrow R$ be defined by:

    Determine if the $\displaystyle \lim_{(x,y) \to (0,0)}$ exists, if it does prove it.

    Here's my attempt:

    As (x,y) -> (0,0) along the y-axis, x=0:

    $\displaystyle \lim_{(x,y) \to (0,0)} y^2 sin(\frac{1}{y^2})$

    Along the x-axis:

    $\displaystyle \lim_{(x,y) \to (0,0)} x^2 sin(\frac{1}{x^2})$

    Along the line y=x:

    $\displaystyle \lim_{(x,y) \to (0,0)} (x^2 +x) sin(\frac{1}{x^2 +x^2})$

    I tried showing whether there is a common limit along different paths but I don't know how to finish this. Any help here is appreciated.
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  2. #2
    MHF Contributor Calculus26's Avatar
    Mar 2009
    You could use the squeezing thm on the 3 limits you have to show all are 0.

    However showing you have a common limit along 3 different paths does not establish the limit. you would have to prove this for every possible path.

    You're method can only be used to show a limit does not exist--i.e. if the limit along different paths is different then the limit does not exist.

    However we from your results we suspect the limit is 0

    So let's return to the definition of limit. I'll use e for epsilon and d for delta

    We need to show

    whenever distance[(x,y) to 0] < d

    that is whenever x^2 + y^2 < d

    then |f(x,y) - L| < e

    |f(x,y) - L| = |f(x,y) - 0|

    =|(x^2 + y^2)sin1/[x^2+y^2] - 0 | < x^2 + y^2

    since |sin1/[x^2+y^2]|<1

    therefore if delta = e then

    |(x^2 + y^2)sin[1/(x^2+y^2)] | < e whenever x^2 + y^2 < d
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