# 2 variable function

#### hyper

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.:)

FU%¤ I posted it in the wrong forum sorry, and I can't find the delete button, please delete this.

### The Physics Forums Way

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving