Is it sufficient to prove that $| f(x,y) – L | = 0 $ so that $\lim_{(x,y)\to(0,0)} f(x,y) =L$ is true? Im trying to proove the following limit by the squeeze theorem:
$$  \lim_{(x,y)\to(0,0)} \frac{xy}{\sqrt{x^2+y^2}} = 0 $$
I've seen all the related questions, and I can't understand the validity of applying absolute value as it follows:
Of course we have that:  $|\frac{xy}{\sqrt{x^2+y^2}}|≥0$ , and also
$ |\frac{xy}{\sqrt{x^2+y^2}}|=  \frac{|x|\cdot |y|}{\sqrt{x^2+y^2}} ≤                    \frac{|x|\cdot |y|}{\sqrt{y^2}} = \frac{|x|\cdot |y|}{|y|}=|x|$, Therfore, by the squeeze theorem:
$$ 0≤|\frac{xy}{\sqrt{x^2+y^2}}|≤|x|  $$
$$ \lim_{(x,y)\to(0,0)}  0≤\lim_{(x,y)\to(0,0)}  |\frac{xy}{\sqrt{x^2+y^2}}|≤\lim_{(x,y)\to(0,0)} |x|  $$
$$ 0≤\lim_{(x,y)\to(0,0)}  |\frac{xy}{\sqrt{x^2+y^2}}|≤0  $$
So finally that should be sufficient to prove that the distance $| f(x,y) – L | = 0 $ and therefore, that should be enough to say that:
$$\lim_{(x,y)\to(0,0)} \frac{xy}{\sqrt{x^2+y^2}} = 0$$
Then my question is: is this valid? Or am I committing a mathematical atrocity.  Thank you for your time and patience.
 A: The problem is that you don't really know if $y = 0$ or not. See, $(x,y)$ needs to not be $(0,0)$ when we're taking the limit but that just means that either $x \neq 0$ or $y \neq 0$. If you do it the way that you did, then you're basically stopping yourself from approaching the origin through any path that goes arbitrarily through the $x$-axis.
Instead, it's probably much better for you to recognize that:
$$|x| \cdot |y| \leq \frac{|x|^2+|y|^2}{2} = \frac{x^2+y^2}{2}$$
Then:
$$\frac{|x|\cdot|y|}{\sqrt{x^2+y^2}} = \frac{1}{2} \sqrt{x^2+y^2}$$
As $(x,y) \to (0,0)$, the right-hand side goes to $0$ without any issues and you obtain the result that you want via the Squeeze Theorem.
Edit:
Right, so what I meant by my paragraph above is the way you reached your derivation of:
$$\frac{|x|\cdot|y|}{\sqrt{x^2+y^2}} \leq |x|$$
There, you just removed the $x^2$ in the denominator. That's going to cause you problems. You could get to the above by recognizing that:
$$\frac{|x|\cdot|y|}{\sqrt{x^2+y^2}} \leq \frac{|x|\sqrt{x^2+y^2}}{\sqrt{x^2+y^2}} = |x|$$
Over here, you don't encounter any problems.
