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Math Help - Taking the limit of this multiple variable function

  1. #1
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    Taking the limit of this multiple variable function

    Hey, hope this is in the right topic

    f(x,y) = (sin(xy)-(xy))/sin(x^2+y^2)

    Thats whats troubling me, I just really dont know where to start and I cant work it out from the solutions.


    Thanks guys.(And gals)
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  2. #2
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    Quote Originally Posted by Monkens View Post
    Hey, hope this is in the right topic

    f(x,y) = (sin(xy)-(xy))/sin(x^2+y^2)

    Thats whats troubling me, I just really dont know where to start and I cant work it out from the solutions.


    Thanks guys.(And gals)
    Which limit are you trying to take? The limit as what tends to what?
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  3. #3
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    Quote Originally Posted by Monkens View Post
    Hey, hope this is in the right topic

    f(x,y) = (sin(xy)-(xy))/sin(x^2+y^2)

    Thats whats troubling me, I just really dont know where to start and I cant work it out from the solutions.


    Thanks guys.(And gals)
     <br />
x \to ? ,y \to ?<br />
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  4. #4
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    I am very sorry, it's at the origin (0,0)
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  5. #5
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    Quote Originally Posted by Monkens View Post
    Hey, hope this is in the right topic

    f(x,y) = (sin(xy)-(xy))/sin(x^2+y^2)

    Thats whats troubling me, I just really dont know where to start and I cant work it out from the solutions.


    Thanks guys.(And gals)
    Let's prove the limit is 0. We know that \sin(u)-u\sim-\frac{u^3}{6} when u\to0. Since xy\to 0 when (x,y)\to(0,0), we have \sin(xy)-xy\sim-\frac{(xy)^3}{6} when (x,y)\to(0,0). Similarly, \sin(x^2+y^2)\sim x^2+y^2, so that f(x,y)\sim-\frac{(xy)^3}{6(x^2+y^2)} when (x,y)\to(0,0).

    Now we can say for instance \frac{|xy|^3}{x^2+y^2}\leq |xy^3|\frac{x^2}{x^2+y^2}\leq |xy^3| (since the ratio is less than 1) hence \frac{(xy)^3}{6(x^2+y^2)}\to0when (x,y)\to(0,0). This concludes.
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  6. #6
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    Thank you very much, that's really helpful I never thought of substituting u in then using the taylor polynomial for that.
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