1. ## Multivariable Limit Problem

I am trying to show that the following function is differentiable at (0,0):

f (x,y) :
((x^4 + y^4) / (x^2 + y^2)) + 1 if (x,y) =/= (0,0)
1 if (x, y) = (0,0)

I have found the Linear Approximation, L at (0 0) = 1
and that R at (0 0) = f (x, y) - L = (x^4 + y^4) / (x^2 + y^2)
and finally, that R / ||(x,y) - (0,0)|| = ((x^4 + y^4) / (x^2 + y^2)) / sqrt(x^2 + y^2)

Now, I need to show that the limit as (x,y) -> (0,0) of R / ||(x,y) - (0,0)|| = 0

Which brings me to my actual problem...I cannot figure out how to go about solving this limit. If anyone can point me in the right direction I would appreciate it.

2. Try putting it in polar coordinates. $x^2+ y^2= r^2$, of course, and $x^4+ y^4= r^4 cos^4(\theta)+ r^4 sin^4(\theta)$ so $\frac{x^4+ y^4}{x^2+ y^2}= r^2\frac{1}{cos^4(\theta)+ sin^4(\theta)}$ and so your "R/||(x,y- (00)||" is $\frac{r^2}{r}\frac{1}{cos^4(\theta)+ sin^4(\theta)}= r\frac{1}{cos^4(\theta)+ sin^4(\theta)}$.

The advantage of polar coordinates here is that the "closeness" to (0, 0) is measured by r alone. No matter what $\theta$ is, that goes to 0 as r goes to 0.

3. I appreciate the help, but I was hoping to figure out how to solve this limit using Squeeze Theorem or Continuity rather than polar coordinates.