Prove that $\lim_{(x,y) \rightarrow (0,0)}\frac{xy}{\sqrt{x^2+y^2}}$ exists without using polar coordinates My only idea to solve this problem was using the definition (epsilon-delta) of limits, but I couldn't come up with a conclusion.
 A: Use that $$x^2+y^2\geq 2|xy|$$ From here we get
$$\frac{xy}{\sqrt{x^2+y^2}}\le \frac{xy}{\sqrt{2|xy|}}$$
and this tends to zero if $$x,y$$ tends to zero.
For $$xy>0$$ we get  $$\frac{1}{\sqrt{2}}(xy)^{1/2}$$ and so on.
A: Note: $2|xy| \le x^2+y^2 $.
$|\dfrac{xy}{\sqrt{x^2+y^2}}| \le $
$(1/2)\dfrac{x^2+y^2}{\sqrt{x^2+y^2}} =$
$(1/2) \sqrt{x^2+y^2}.$
Choose $\delta =2\epsilon$.
A: Note that if $\sqrt{x^2+y^2}<\delta$ so both of $|x|$ and $|y|$ are less than $\delta$. I f-for example- take $z=\text{max}\{|x|,|y|\}$ so $z>=0$ and $$\sqrt{x^2+y^2}>|\sqrt{z+0}|$$ This can lead you t get a right proof. Here, the possible limit is zero. In fact, having a path-wise proofing make us sure that the limit is zero, so we should proof it by $\epsilon-\delta$ version.
A: $$\left|\dfrac{xy}{\sqrt{x^2+y^2}}\right|<\dfrac{|xy|}{\sqrt{2|xy|}}<\sqrt{|xy|}$$
A: $$|x| \leq \|(x, y) \|=\sqrt{x^2+y^2}$$
So:
$$0 \leq \frac{|x||y|}{\sqrt{x^2+y^2}} \leq |y|$$
And $$\lim_{(x,y) \rightarrow (0,0)}|y|=0$$
So by the squeeze theorem you're done.
