[Not to be turned in]
Consider the function $\displaystyle f(x,y) = {x^2 y\over x^4 + y^2}$.
Note that this function is not defined at $(x,y) = (0,0)$ because there is division by zero. But does it have a limit there? We saw an example of a function in class where that wasn't the case, because the limit along any line toward the origin gave a different value.
For this funciton, the situation is different. The limit along any straight line toward the origin is 0, which would suggest that the limit should be 0. But if you take a path along any parabola with vertex at the origin, the limit is not 0.
Your goal, here, is to show these two facts.
Because the limits along the parabolas don't match, the function doesn't have a limit as $(x,y)$ goes to $(0,0)$, even though along any striaght line, the limit was zero.
 Consider the line $y=mx$ for any constant $m$, which is a line through the origin with slope $m$. Then along this line, the value of the function is $z=f(x,mx)$. Show that the limit of this value as $x$ goes to $0$ is $0$ for every $m$.
(Note: you will have to treat the case $m=0$ as a special case.)
 In the previous part, you handled every line through the origin except the vertical line, $x=0$. The values of $f$ along this line are $z=f(0,y)$. Show that the limit of this value as $y$ goes to $0$ along this line is also $0$.
 This would suggest that the limit at $(0,0)$ should be $0$, but it turns out this isn't the case. Consider the parabola $y=ax^2$ for a constant value $a$. The values of $f$ along this parabola are given by $z=f(x,ax^2)$. Show that, for any $a$, these values are constant along the parabola, and that the constant is not zero unless $a = 0$.
We will look at the graph of this function on Friday to see what is going on.
I am also putting up a new WeBWorK assignment due next Monday. Some of it will rely on Friday's lecture, so don't be concerned if you can't do it all just yet.

