Proving continuity in the origin $\frac{xy}{\sqrt{x^2 + y^2}}$

Let $$g: \mathbb{R^2} \to \mathbb{R}$$.

How can I prove that $$g$$ is continuous in its origin, but not totally differentiable?

If I take

$$g(\frac{1}{n},\frac{1}{n}) = \frac{1}{n\sqrt{2}} \to 0 \text{ for } n \to \infty$$

Or rather:

$$|x,y| \leq \frac{1}{2} (x^2 + y^2)$$

from which we can follow

$$|g(x,y)| \leq \frac{1}{2} \sqrt{x^2 + y^2}$$

which proves continuity of $$g$$ in the origin $$(0,0)$$.

But how can I show that this function is not total differentiable?

Can I do the following estimation?

$$|g(x,y) - 0| = |y| \cdot \frac{x \cdot y}{\sqrt{x^2 + y^2}} \leq |y| \cdot 1 = |y|$$

From which it follows, that

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

• – Yagger Nov 8 '18 at 22:01

By polar coordinates as $$r \to 0$$

$$\frac{xy}{\sqrt{x^2 + y^2}}=r \cos \theta \sin \theta$$

we see that $$f(x,y)$$ is continuous but since $$f_x(0,0)=f_y(0,0)=0$$ by definition of differential we have

$$\lim_{(h,k)\to (0,0)} \frac{\frac{hk}{\sqrt{h^2 + k^2}}}{\sqrt{h^2 + k^2}}=\frac{hk}{h^2 + k^2}$$

which doesn't exist.

Refer also to the related

Note that $$x^2+y^2\ge 2|xy|$$. Hence, we have

$$\left|\frac{xy}{\sqrt{x^2+y^2}}\right| \le \frac{\sqrt{|xy|}}{\sqrt 2}$$