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anonymous
 4 years ago
I need to show the following limit exists. lim(x,y) > (0,0) of xy(xy)/(x^2 + y^2). I can probably plug in for y=0 and x=0, but how can I show the limit exists for all lines?
anonymous
 4 years ago
I need to show the following limit exists. lim(x,y) > (0,0) of xy(xy)/(x^2 + y^2). I can probably plug in for y=0 and x=0, but how can I show the limit exists for all lines?

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Try using polar coordinates: \[x = r\cos(\theta)\]\[y = r\sin(\theta)\] Now your limit becomes: \[\lim_{(x,y) \rightarrow (0,0)} [x \times y \times (xy)] \div [x^{2} + y^{2}] = \] \[\lim_{r \rightarrow 0} [r \cos(\theta)\times r \sin(\theta)\times (r \cos(\theta)  r \sin(\theta))]\div[r^2 \cos^2(\theta) + r^2 \sin^2(\theta)] \] \[\lim_{r \rightarrow 0} [r^2\cos(\theta)\sin(\theta)\times r(\cos(\theta)  \sin(\theta))]\div r^2\]Now, observe that it's a limit of "r" going to 0, so r can never be 0, therefore we can divide by r² the top and the bottom our ratio:\[\lim_{r \rightarrow 0} r \times \cos(\theta)\sin(\theta)\times(\cos(\theta) \sin(\theta)) = \]\[\cos(\theta)\sin(\theta)\times(\cos(\theta) \sin(\theta))\lim_{r \rightarrow 0}r = 0\]Meaning that, independently of the angle (the theta) you are approaching the origin, it will always go to zero. Hope it helped!
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