Differentiability of $f(x, y) = |xy|$ at $0$; is this a mistake by Munkres?

Solution 1:

We have that

$$\frac{x^2y^2}{x^2+y^2}\leq x^2+y^2,$$

so if $||(x,y)||=r$ then

$$\frac{|xy|}{||(x,y)||}=\sqrt{\frac{x^2y^2}{x^2+y^2}}\leq r,$$

and as such

$$\lim_{x,y\to 0} \frac{|xy|}{||(x,y)||}=0$$

as $(x,y)\to(0,0)$ implies that $||(x,y)||\to 0$. The reason your statement does not imply nondifferentiability is that the fact that

$$D_1f(x, y) = \begin{cases} |y|&\text{if}\, x>0 \\ -|y| &\text{if}\, x<0 \end{cases} $$

is only problematic when $y$ is away from $0$, otherwise $|y|\approx-|y|$. The partials still very much exist at $f(0,0)$, as $|y|,-|y|\to 0$. However, you can use your result to conclude that $D_1 f(x,y)$ does not exist if $y\neq 0$.

Solution 2:

How did you conclude that $D_1f(0,0)$ does not exist? Try using the definition directly. You are probably confusing differentiable and continuously differentiable.