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Let $f(x,y)=0$ if $y\leq0$ or if y$\geq x^2$ and let $f(x,y)=1$ if $0<y<x^2$. Show that $f(x,y) \rightarrow 0$ as $(x,y) \rightarrow (0,0)$ along any straight line through the origin. Find a curve through the origin along which (except at the origin) $f(x,y)$ has the constant value $1$. Is $f$ continuous at the origin?

Thank you

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It would help if you posted what have you tried and where you are confused, so we can provide better guidance. Regards – Amzoti Mar 10 '13 at 0:47
I might help to draw $y=x^2$ and to locate the regions where $y\geq x^2$ and $0<y<x^2$. – 1015 Mar 10 '13 at 0:56
up vote 1 down vote accepted

Here are some vague hints. So you have $$ f(x,y) = \begin{cases} 0 & \text{if } y\leq 0 \\ 0 & \text{if } y\geq x^2 \\ 1 & \text{if } 0< y < x^2\end{cases} $$ Let $y = ax$ be a straight line through the origin. Lets just say that $a > 0$. You want to show that along this line the limit is $0$. Well, clearly on the part of the line where $y\leq 0$, the function is $0$, so you only have to consider $y > 0$. Now (you can write down the details yourself) for small enough $x$, you will have $x^2 < ax$, and so when you consider the limit as $x\to 0^+$ you get $$ \lim_{x\to 0} f(x,ax) = \lim_{x\to 0} 0 = 0 $$ What curve might you try to get a limit that isn't equal to zero? How about $y = \frac{1}{2}x^2$? Along the curve you will have the limit as $x\to 0$ equal to ... (I will let you insert the last details.)

Now, go look up the definition of continuous and ask yourself the question of whether the function can be continuous when you have found two curves along which the limits are different.

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I don't know how to continue. I tried. I really did – Carpediem Mar 10 '13 at 2:00

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