Stephen M. answered • 12/28/16

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f(x,y) = 1/(x

^{2}- y)(a) For critical points, we need to find the points where both partial derivatives equal zero. So, using chain rule,

δf/δx = -1 / (x

^{2}- y)^{2}* 2xSetting this equal to zero, we find x = 0

δf/δy = -1 / (x

^{2}- y)^{2}* (-1)Setting this equal to zero, there are no solutions. Therefore, this function has no critical points. We would need to find a point where both partial derivatives are simultaneously zero to have a critical point.

(b) Conceptually, a limit of a multivariable function existing means we can approach that point from any direction (or path) and get the same limit. In this case, consider simply the line x = 0 and decide for yourself whether the limit is the same as y approaches 0 from the positive and negative side. Demonstrating that the limit from two different directions (or along two different paths) is different is sufficient to prove a limit doesn't exist, just as it is for single variable functions.

If you need clarification on that, feel free to let me know.

Stephen M.

Exactly. Although each axis can be approached from either direction. So for the y axis (x=0), we get a negative approaching from y>0 and a positive approaching from y<0. But yes, it's also positive along the x-axis y=0 (in both directions).

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12/28/16

Claire H.

12/28/16