Can the function $f(x,y) = \frac{xy}{\sqrt{x^2+y^2}}$ be defined so that $f$ is continuous and differentiable at the origin?

I redefined the function piecewise so that $f=0$ at the origin and $f = \frac{xy}{\sqrt{x^2+y^2}}$ otherwise. This made the function continuous since $\displaystyle\lim_{(x,y) \to (0,0) } \frac{xy}{\sqrt{x^2+y^2}} = 0$.

However, I'm having a difficult time of proving whether or not f is differentiable. I'm pretty sure it's not. I tried to do this by showing that the partial derivatives are not continuous at the origin, but I got stuck.



Let us check whether $\frac{\partial f}{\partial x}(0,0)$ exists. We have

$$ \frac{\partial f}{\partial x}(0,0) = \lim_{h \to 0} \frac{f(h,0) - f(0,0)}{h} = \lim_{h \to 0} 0 = 0 $$

and similarly, since the function is symmetric in $x$ and $y$, $\frac{\partial f}{\partial y}(0,0) = 0$ so both partial derivatives exist.

Those are necessary but not sufficient conditions for the function $f$ to be differentiable at the origin. For $f$ to be differentiable at the origin, we must have

$$ \lim_{(x,y) \rightarrow 0} \frac{f(x,y) - f(0,0) - \frac{\partial f}{\partial x}(0,0) x - \frac{\partial f}{\partial y}(0,0) y}{||(x,y)||} = 0. $$

Plugging in, we have

$$ \frac{f(x,y) - f(0,0) - \frac{\partial f}{\partial x}(0,0) x - \frac{\partial f}{\partial y}(0,0) y}{||(x,y)||} = \frac{xy}{x^2 + y^2}. $$

Denoting this function by $g(x,y)$, we have

$$ \lim_{t \to 0} g(t,t) = \lim_{t \to 0} \frac{t^2}{t^2 + t^2} = \lim_{t \to 0} \frac{1}{2} = \frac{1}{2}, \,\,\, \lim_{t \to 0} g(0,t) = \lim_{t \to 0} 0 = 0 $$

so $ \lim_{(x,y) \to (0,0)} g(x,y)$ doesn't exist and $f$ is not differentiable at the origin.


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