# Limit of a multivariable function

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

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.)