- #1

- 171

- 0

## Homework Statement

could you please check if this exercise is correct?

thank you very much :)

##f(x,y)=\frac{ |x|^θ y}{x^2+y^4}## if ##x \neq 0##

##f(x,y)=0## if ##x=0##

where ##θ > 0## is a constant

study continuity and differentiabilty of this function

## The Attempt at a Solution

Being the quotient of continuous functions, it is always continuos except, maybe, in the point (0,0)

I normally use the sequential continuity to tell is a function of this kind is continuous. However, I couldn't find a couple of suitable successions for this case, so I decided to proceed as follows:

## \left| \frac{|x|^θ y}{x^2+y^4} - f (0,0) \right| ≤ \left| |x|^θ y \right|##

that goes to 0 if x,y go to 0. So the function is continuous in (0,0).

Then it could be differentiable. I look for the partials:

##\frac{\partial f}{\partial x}\lim_{t \to 0} \frac{f(x_0+t)-f(0,0)}{t}=0##

and

##\frac{\partial f}{\partial y}=0## as well. if they are continuous in 0,0 the function is differentiable for the theorem of total differential. but the partials ar every "complicated", so i prefer to apply the definition of differential:

##\lim_{(x,y) \to (0,0)} \frac{f(x,y)-f(0,0)-(∇f(0,0),(x,y)}{\sqrt{x^2+y^2}}##

=## \left| \frac{|x|^θ y}{(x^2+y^4) \sqrt{x^2+y^2}}-0 \right| ## but i see it is my original function multiplied ##\frac{1}{\sqrt{x^2+y^2}}##. this last function goes to infinity for x,y going to 0,0, so the limit doesn't exist and the function is not differentiable.