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Math Help - Two-Variable Limit and Continuity

  1. #1
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    Two-Variable Limit and Continuity

    I'm having trouble with this problem

    f(x,y)= {(sin(x^2+y^2 )/((x^2+y^2 ) ) (x,y)=(0,0)
    1 (x,y)≠(0,0)


    The question says to show that f is continuous
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  2. #2
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    Quote Originally Posted by ghow90 View Post
    I'm having trouble with this problem

    f(x,y)= {(sin(x^2+y^2 )/((x^2+y^2 ) ) (x,y)=(0,0)
    1 (x,y)≠(0,0)


    The question says to show that f is continuous
    I assume that you actually mean

    f(x, y) = \frac{\sin{(x^2 + y^2)}}{x^2 + y^2} for (x, y) \neq (0, 0)

    and f(x, y) = 1 for (x, y) = (0, 0).


    To show that this is continuous, you'd have to show that

    \lim_{(x, y) \to (0, 0)} f(x, y) = 1.


    \lim_{(x, y) \to (0, 0)} f(x, y) = \lim_{(x, y) \to (0, 0)} \frac{\sin{(x^2 + y^2)}}{x^2 + y^2}

     = \lim_{r \to 0} \frac{\sin{r^2}}{r^2}

    since r^2 = x^2 + y^2.



    It is well known that \lim_{X \to 0} \frac{\sin{X}}{X} = 1 (and can be shown using the sandwich theorem).


    Therefore \lim_{r \to 0} \frac{\sin{r^2}}{r^2} = 1.


    Since this limit is 1, the function is continuous.
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  3. #3
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    Quote Originally Posted by Prove It View Post
    I assume that you actually mean

    f(x, y) = \frac{\sin{(x^2 + y^2)}}{x^2 + y^2} for (x, y) \neq (0, 0)

    and f(x, y) = 1 for (x, y) = (0, 0).


    To show that this is continuous, you'd have to show that

    \lim_{(x, y) \to (0, 0)} f(x, y) = 1.


    \lim_{(x, y) \to (0, 0)} f(x, y) = \lim_{(x, y) \to (0, 0)} \frac{\sin{(x^2 + y^2)}}{x^2 + y^2}

     = \lim_{r \to 0} \frac{\sin{r^2}}{r^2}

    since r^2 = x^2 + y^2.



    It is well known that \lim_{X \to 0} \frac{\sin{X}}{X} = 1 (and can be shown using the sandwich theorem).


    Therefore \lim_{r \to 0} \frac{\sin{r^2}}{r^2} = 1.


    Since this limit is 1, the function is continuous.





    Thank You for the help
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  4. #4
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    Notice that Prove It changed to polar coordinates. That simplifies the problem greatly- since just r itself measures the distance to (0,0) regardless of \theta. As long as the limit, as r goes to 0, does NOT depend on [tex]\theta[\math], that gives the correct limit.
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