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Find an example of a function $ f: \mathbb{R^2} \to \mathbb{R} $ so that $f$ is not continuous at $(0,0) \in \mathbb{R}$, but $ \lim_{t \to 0} f(tx) = 0$ where $tx= (tx_1, tx_2)$ (so $t \in \mathbb{R} $ and $x \in \mathbb{R^2}$). How can I tell my example works? So I chose $f: \mathbb{R}^2 \to \mathbb{R}$, $f(x,y)= \frac{x^2y}{x^2+y^2}$ since it is not continuous at $(0,0)$, but then I don't know how to prove that $ \lim_{t \to 0} f(tx) = 0$.

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    Luz: That isn't correct. Did you use $y^2$ in the numerator instead of $y$?2012-11-28

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You must specified what is $f(0,0)$. To be discontinuous at $(0,0)$ you must define $f(0,0)\neq 0$ because $\displaystyle{\lim_{(x_1,x_2)\to(0,0)}f(x_1,x_2)=0.}$ Can you see why?(Hint: $|f(x_1,x_2)|\leq |x_2|$).
Since $\displaystyle{\lim_{(x_1,x_2)\to(0,0)}f(x_1,x_2)=0}$ you obtain $\displaystyle{\lim_{t\to 0}f(tx_1,tx_2)=\lim_{(x_1,x_2)\to(0,0)}f(x_1,x_2)=0}$.
Or just note that for $(x_1,x_2)\in \mathbb{R}^2-\{(0,0)\}, \ \ f(tx_1,tx_2)=t\dfrac{x_1^2x_2}{x_1^2+x_2^2}$.
For more examples take a continuous function $g$ at $(0,0)$ with $g(0,0)=0$ and change its value at $(0,0)$.
If you want an example with 'non-removable' discontinuity at $(0,0)$ Jonas Meyer's example at comments is better.

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    Thank you. I deleted my erroneous comment. I had been thinking of a different "standard" example where you can set $y=x^2$ to get a nonzero nonradial limit, and clearly did not think here (about the limit, that is).2012-11-28