# Function with 2 variables

This is the problem, we have:

f(0,0)=0 for (x,y) not(0,0) f(x,y)= $$\frac{2*x^3*y}{x^6+y^2}$$

a) What is lim f(x,y) (x,y)=->(0,0) alonng any straign line.

b) Show that in every circle centered at the origin there is a point where f(x,y)=1

c) Decide if f(x,y) is continious at the origin.

d)Decide if the partial derivatives at the origin exicsts, and if the do what are they?

my attempt at the sollution:

a) Here I used polarcoordinates and found out that the limit valuse was zero, so I dont need you to try on this.

b) If you choose y=x^3 it is allways one there, so I think I got this, but I want you more to look at the last 2 questions.

c) Now comes the tricky part. Along every straight line the limit valuse is zero, and f(0,0)=0 so it might excist. But if we try the let y have the value y=k*x^3, f(x,y) is allways $$\frac{2k}{1+k^2}$$ and it should also have this value when it goes to the origin, so so if we choose k not 0 we dont get the limit zero and it isnt contious. Do you guys have any ideas here?

d) Here I tried partial differensiation with the expression and evaluating in (0,0), but you can't do this because the you would divide with zero, and that is not ok to do. So maybe I should combine f(0,0)=0 and f(x,y)= $$\frac{2*x^3*y}{x^6+y^2}$$ ? Maybe the partial derivative do not excist there continious?

Help would be greatly appreciated.:)