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Common approach in handling functions of two variables is to express this function in polar coordinate system. For example, in the classic example $$f(x,y)=\left\{\begin{array}{lr}\frac{xy}{x^2+y^2} & (x,y)\neq (0,0)\\0 & \text{otherwise}\end{array}\right.$$ say $x=r\cos(\phi)$ and $y=\sin(\phi)$, and then we can write $f(r, \phi)=\frac{r^2\sin(\phi)\cos(\phi)}{r^2(\cos^2(\phi) + \sin^2(\phi))}=\frac{1}{2}\sin(2\phi)$. Now if we let $r\rightarrow 0$, then it and, for example, $\phi=\frac{\pi}{4}$, the $$\lim_{r\rightarrow 0}f(r, \frac{\pi}{4})=\frac{1}{2}\neq 0$$ so $f$ isn't continuous in $(0,0)$.

Now I am given a function $$f(x,y)=(x+y)\exp(xy)$$Since this function is composition of continuous functions, it must be continuous, but I wanted to ask in common: earlier I used polar coordinates to tell that the function isn't continuous, so can I use the same approach to say that function is continuous, that is: $$\lim_{r\rightarrow0}f(r,\phi)=\lim_{r\rightarrow0}r(\cos(\phi)+\sin(\phi))\exp(r^2\sin(\phi)\cos(\phi))=0$$ so function must be continuous, or is it not enough?

Thank you in advance!

nakajuice
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Yes, that should be enough. As long as you are sure to take an arbitrary $\phi(r)$ (continuous, of course) for your angle (which will serve to parameterize the path in $\Bbb R^2$ towards the point), you can just take the limit as $r\to 0$ to see if it depends on $\phi(r)$. If not, then there is no dependence on path: i.e. the limit exists, and your function is continuous. Be careful though, sometimes the dependence on the path is subtle (watch out for trig functions in denominators)... However, here it looks like there's no problem: there's nothing to stop that $r$ in front from making everything $0$.

Stahl
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