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How can I calculate this limit?

$$\lim _{(x,y) \to (0,0)} \frac{\vert{x\vert\vert{y}\vert}}{x^2 +y^2}$$

I don't have idea and I will be appreciate for your help.

user
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Mr.Yoon
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2 Answers2

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Suppose $\;x=0\;$ , then the limit is clearly zero, but if $\;y=x\;$ then

$$\frac{|x||y|}{x^2+y^2}=\frac{x^2}{2x^2}=\frac12\xrightarrow[x\to0]{}\frac12$$

Thus, the limit doesn't exist.

DonAntonio
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1

Go polar to obtain

$$\frac{\vert{x\vert\vert{y}\vert}}{x^2 +y^2}=\vert{\cos \theta\vert\vert{\sin \theta}\vert}$$

user
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