Analyzing the continuity at a point of $f:\mathbb{R} ^{2}\rightarrow \mathbb{R}$

So i have the following function, $f:\mathbb{R} ^{2}\rightarrow \mathbb{R}$, and they ask me to analyze the continuity at the point $P = (1,0)$, when f is defined as follows:

$$f(x,y) =\begin{cases} \dfrac{\left( x^{2}-2x+1\right) y\cos \left( \dfrac {1} {\left( x-1\right) ^{2}+y^{2}}\right) } {\left( 3\left( x-1\right) ^{2}+y^{2}\right) \left( 2+x^{2}-2x\right) } & \text{ if }(x,y) \neq (1,0)\\ 0 &\text{ if } (x,y) = (1,0)\end{cases}$$

So, I know that, in order to be continuous, the limit of $f$ at $(1,0)$ has to be $0$ (the value of the function at the given point). But i don't know how to prove that it's (or it isn't) $0$.

Can you anyone give me a hint of how to face these problem? If you have, any link with similar problems where i can learn the method? Because the type of exercise are very frequent in the exams, but are not developed in the textbook we use (Mardsen and Tromba, Vol. 3).

Thanks a lot for your help!

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A function $f(x,y)$ is continuous on a line $l(t) = (x(t),y(t))$ if $f \circ l$ is continuous. First check that this function is continuous on any line through the point $(1,0)$ (a relatively weak condition). I suspect it will be. However, I bet you can find a parabola through $(1,0)$ on which $f$ is not continuous. That is just the "smell" i get from having seen similar problems before. – orlandpm Dec 7 '12 at 1:21

1 Answer

Simplify your expression by a change of variable $(x-1,y)=(r\cos\theta,r\sin\theta)$ with $r\ge0$ and $\theta\in[0,2\pi)$. That is, first translate the origin to $(1,0)$, and then use polar coordinates. Then show that $f\rightarrow0$ as $r\rightarrow0$.

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any other ideas? I'm not familiar with polar coordinates (i should! but in these course we don't use them :S). – pmartelletti Dec 7 '12 at 3:50
If you're not familiar with polar coordinates, get used to it. It's not that difficult. For instance, by completing square, the $2+x^2-2x$ in the denominator becomes $(x-1)^2+1$. Hence it can be simpified to $r^2\cos^2\theta+1$. – user1551 Dec 7 '12 at 14:48