# Determining whether a function of two variables is continuously differentiable

I'm just trying to solidify the different cases for functions When dealing with functions that are $C^1$, say I have some function: $$f(a,b)=\begin{cases}(a^4+b^4)/(a^2+b^2)\quad &\text{ if } (a,b)\ne (0,0)\\ 0 &\text{ at }(0,0)\end{cases}$$

$C^1$ functions need to have their partials exist and be continuous everywhere in that neighbourhood. So to test if my function is continuously differentiable, I can first take the classical limit of the formula and test to see that the partials exist (that is, by fixing $y$, and adding arbitrary $h$ to $x$, and taking limit as $h\to0$ and vice versa for $y$ with $k\to0$).

Since they do in this case, do I then need to test if the partials $=0$ as $(a,b)\to(0,0)$ (taking the limit of the partials). Or would I take the limit of the function itself?

And if this function is continuously differentiable (which the limit does go to $0$, in both the original $f$, and $f'$ -- I'm not sure which one is more important), this function is differentiable everywhere, and thus $C^1$ ?

To check that such a function is $C^1$, you should

1. Show that $\lim_{h\to 0}\frac{f(h,0)-f(0,0)}{h}$ and $\lim_{h\to 0}\frac{f(0,h)-f(0,0)}{h}$ exist. These are the partial derivatives at $(0,0)$; they must be computed by the definition because the usual limit theorems do not apply at $(0,0)$.

2. Compute partial derivatives at points $(x,y)$ other than $(0,0)$. This is an exercise with the quotient rule.

3. Show that the result of step 2 tends to the result of step 1 as $(x,y)\to (0,0)$.

You do not need to verify that $f$ itself is continuous, since you can invoke a theorem that says that continuous partials imply differentiability, and therefore continuity.

On the other hand, sometimes a function of this type turns out to be discontinuous at $(0,0)$, and then it's good idea to demonstrate that, thus showing that it's not $C^1$.

• Thank you kindly. Clear and concise, just what I needed to sort the textbook's dense (although useful) and convoluted info. – sahimat Oct 3 '14 at 5:02