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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.

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    Some related posts: [show continuity of $\frac{xy}{\sqrt{x^2+y^2}}$](http://math.stackexchange.com/questions/572098/show-continuity-of-fracxy-sqrtx2y2) and [Finding partial derivative of $f(x,y)=\frac{xy}{\sqrt{x^2+y^2}}$](http://math.stackexchange.com/questions/1422391/finding-partial-derivative-of-fx-y-fracxy-sqrtx2y2).2015-09-27

1 Answers 1

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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?