Prove that $\lim_{(x,y) \rightarrow (0,0)}\frac{xy}{\sqrt{x^2+y^2}}$ exists without using polar coordinates

Question

My only idea to solve this problem was using the definition (epsilon-delta) of limits, but I couldn't come up with a conclusion.

Answer 1

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

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.

Answer 2

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.

Answer 3

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.

Answer 4

$$\left|\dfrac{xy}{\sqrt{x^2+y^2}}\right|<\dfrac{|xy|}{\sqrt{2|xy|}}<\sqrt{|xy|}$$

Answer 5

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 $\delta =2\epsilon$.