f(x,y) = (x3+y3)^(1/3)
Show that fy(0,0) = 1
The Attempt at a Solution
And...I take the limit of it as x and y goes to zero, which gets me 0/0
If the limit exists (we don't yet know if it does, but the question posed seems to assume it does), then you can evaluate it by choosing a "path" to "lim" along.
The first part of that isn't true- you don't "have to do it along some path". The best way to evaluate limits in two dimensions, as (x, y) goes to (0, 0) is to change to polar coordinates. That way, r alone measures the distance to (0, 0). If the limit, as r goes to 0, does not depend on [itex]\theta[/itex], then the limit exists and is equal to that value.When you take the limit of a multivariable function, you have to do it along some path [itex]y(x)[/itex]...If every path leads to the same result, then the limit exists and is equal to that result.