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Mathematics 8 Online
OpenStudy (anonymous):

Evaluate: lim (x,y)->(0,0) of (xy^2)/(x^2+y^2)^(3/2)

OpenStudy (anonymous):

\[\lim_{(x,y)\rightarrow 0} \frac{ xy^2 }{ (x^2+y^2)^{3/2} }\] Do you mean this?

OpenStudy (zarkon):

I'm sure he means \[\lim_{(x,y)\rightarrow (0,0)} \frac{ xy^2 }{ (x^2+y^2)^{3/2} }\]

OpenStudy (anonymous):

@Zarkon Correct! Just as a side note how do you write equations when asking a question (I know how to do it in replies)

OpenStudy (anonymous):

I think it works the same way in the questions as it does in replies, it's just that replies doesn't give you an equation editor.

OpenStudy (anonymous):

i am still in single var i'll try \[\frac{ xy^2 }{ (r^2) ^{3/2}}\] \[\frac{ xy^2 }{ r^3 }=\frac{ r^3\sin ^2\theta \cos \theta }{ r^3}=\cos \theta-\cos^3\theta\]

OpenStudy (anonymous):

1-1^3=0

OpenStudy (anonymous):

\[r \cos \theta \rightarrow 0\]

OpenStudy (zarkon):

are you assuming \(\theta=0\)?

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