$f: \Bbb R^2 \to \Bbb R$ defined by $f(x,y)=\frac {xy^2}{x^2+y^2}$ if $(x,y)$ not equal $0$, $o$ if $(x,y)=0.$ Is this function continuous at $(0,0)$?

Question

So, what I was thinking is that you could change this into polar coordinates so that $f(r,\theta)=\frac {(r^3\cos \theta \sin^2\theta)}{r^2}$ so then the limit as $(r,\theta)$ approaches $(0,0)$ is $0$. So then it would be have a limit at zero and therefore be continuous at zero.

But I know that the examples in the book use just the definition of continuity, and I didn't know if I am supposed to use that as well, and how it would apply to this one. (For example they showed that the same question except substituted $f(x,y)=\frac{xy}{\sqrt{x^2+y^2}}$ when $(x,y)$ does not equal zero and $0$ if $(x,y)=0$ is continuous at $(0,0)$ just using the definition of continuity.

$f(x,y) = \frac{x}{\frac{x^{2}}{y^{2}}+1}$, when $(x,y)\neq (0,0)$ (just cancel out $y^{2}$). Now, use that $\frac{x}{\frac{x^{2}}{y^{2}}+1} \leq \frac{x}{1} = x \to 0$, when $(x,y)\to (0,0)$. — 2017-02-05

user id:413489 6 · Accepted Answer

Oh, or could I just use $|f(x,y)-f(0,0)|=|{xy^2\over x^2+y^2}|=|{xy\over x^2+y^2}||y|\le {1\over 2}|y|$ and then take $\delta=2\epsilon$ ?

$f: \Bbb R^2 \to \Bbb R$ defined by $f(x,y)=\frac {xy^2}{x^2+y^2}$ if $(x,y)$ not equal $0$, $o$ if $(x,y)=0.$ Is this function continuous at $(0,0)$?

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