The following limit exists, but I can't find its value: $\displaystyle \displaystyle\lim_{(x,y)\to(0,2)}\frac{(y-2)^2\sin(xy)}{x^2+y^2-4y+4}.$

Hope you can help me. :D

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- Feb 7th 2011, 12:10 PMKillerA two variable limit
The following limit exists, but I can't find its value: $\displaystyle \displaystyle\lim_{(x,y)\to(0,2)}\frac{(y-2)^2\sin(xy)}{x^2+y^2-4y+4}.$

Hope you can help me. :D - Feb 7th 2011, 01:54 PMtopsquark
If the limit exists (and I haven't gone so far as to prove that) then it must exist for x, y --> (0, 2) from any way of approaching the point (0, 2). What I would do is to take, say, y = x + 2. Then after you sub y = x + 2 into the limit, you evaluate it for x --> 0.

-Dan - Feb 7th 2011, 02:22 PMGeneral
$\displaystyle \left| \dfrac{(y-2)^2 sin(xy)}{x^2+y^2-4y+4} \right| \leq sin(xy)$

Use sandwich.