Let $f(x,y) = \dfrac{4xy(x+y)}{(x^2+y^2)} $, $f(0,0)=0$.
Show that if $f(x, y)$ is differentiable at $(0, 0)$, then $\lim _{(h,h)\rightarrow (0,0)}f(h,h) = 0$.
Finally, show that $\lim _{(h,h)\rightarrow (0,0)}f(h,h)\ne0$ (does not equal to zero)
I used the limit definition to show that $f(x,y)$ is differentiable and equal to $0$. If I show that the partial derivatives with respect to $x$ and $y$ exist and are $0$, shouldn't the $\lim _{(h,h)\rightarrow (0,0)}f(h,h)$ also be $0$ for $2$)?
edit: $f(x,y)$ is not differentiable at $(0,0)$, but the partial derivatives with respect to $x$ and $y$ do exist and are equal to $0$. Sorry for the confusion
$$(1.)$$Assume that $f(0,0)$ exists.If $f$ is differentiable at $(0,0)$ then $f$ is continuous at $(0,0)$ .If $f$ is continuous at $(0,0)$ then $f(0,0)=\lim_{n \to \infty}f(1/n,1/n)=\lim_{n \to \infty}4/n=0.$ If $f(0,0)=0$ then $f$ IS continuous at $(0,0)$, for if $0<x^2+y^2<r$ then $|x y (x+y)|\le |x^2 y|+|y^2 x|\le (x^2 +y^2) \max(|x|,|y|)\le (x^2+y^2)r$, so $|f(x,y)|\le 4[(x^2+y^2)r]/[x^2+y^2]=4r$. In particular $f(h,h)$ DOES converge to $0$ as $h\to 0$ in non-zero values of $h$ regardless of the existence of $f(0,0)$. $$(2.)$$ Do not assume that $f(0,0)$ exists. Then nothing is known about whether or not $f$ is continuous or differentiable at $(0,0).$ $$(3.)$$ Assume that $f$ takes the value given by the initial formula for every $(x,y)$ in its domain.Then $f(0,0)$ does not exist, and so $f$ cannot be continuous or differentiable or anything else at $(0,0)$.......I think this was the point :Don't make unjustifiable assumptions, no matter how small.If $f$ was not defined by me, I can't add extra info into the def'n and still say that it's the same $f$.