# Continuity and differentiability of the function $f(x,y)=\frac{xy}{\sqrt{x^2+y^2}}$

Would like some guidance. What I've done so far is included.

Given,

$$f(x,y)=\begin{cases} 0, \text{ if } (x,y)=(0,0)\\ \\ \frac{xy}{\sqrt{x^2+y^2}}, \text{ if } (x,y)\ne (0,0) \end{cases}$$

Prove

a. $f$ is continuous (at all points)

Find a function that bounds f. Take the limit, this should show this. I can't find a bounding function.

b. $f$ has partial derivatives (at all points)

I take the $\partial_u f= \nabla f \cdot u$. I use $u = \sin x, \cos x$. I've got $f = \cos x \sin x$. but am not sure what to do it.

c. $f$ is not differentiable at $(0,0)$.

Show that the limit approaches two different points. I can't find paths such that.

• This is quite easy, what did you try? – Jonas Teuwen Oct 24 '11 at 8:43
• i've edited in some comments on what i've tried. – user18094 Oct 24 '11 at 8:50
• For (a), try polar coordinates. (That might help for (c) as well.) – Hans Lundmark Oct 24 '11 at 9:12
• – Did Oct 24 '11 at 10:35
• – Martin Sleziak Sep 27 '15 at 5:38

For (a) you can follow Hans Lundmark’s suggestion. Alternatively, you can use the fact that $$0\le (x-y)^2 = x^2-2xy+y^2$$ to deduce that $2xy\le x^2+y^2$, from which it’s easy to bound the function.
For (c), what happens if you approach the origin along the line $y=x$? Then for $x\ne 0$ your function is just $$f(x) =\frac{x^2}{\sqrt{2x^2}},$$ whose derivative is pretty easy to investigate. What if you approach the origin along the $x$-axis?