Can I calculate limits in 3 dimensions with a substitution of the type $z=xy$? This is my first question here so I hope I'm doing it right :) sorry otherwise!
As in the title, I was wondering if and when it is OK to calculate a limit i three dimensions through a substitution that "brings it down to two dimensions". Let me explain what I mean in a clearer way through an example. I was calculating this limit:
$$\lim_{(x,y) \to (0,0)} \frac{\ln (1+\sin^2(xy))}{x^2+y^2} =\lim_{(x,y) \to (0,0)} \frac{\ln (1+\sin(xy)\cdot \sin(xy))}{x^2+y^2}$$
$$=\lim_{(x,y) \to (0,0)} \frac{\ln (1+xy\cdot xy)}{x^2+y^2} =\lim_{(x,y) \to (0,0)} \frac{\ln (1+x^2y^2)}{x^2+y^2}=\lim_{(x,y) \to (0,0)} \frac{x^2y^2}{x^2+y^2}$$
$$=\lim_{(x,y) \to (0,0)}\frac{1}{\frac{1}{y^2}+\frac{1}{x^2}}="\frac{1}{\infty}"=0.$$
Where I have used:
$$ \lim_{(x,y) \to (0,0)} \frac{\sin(xy)}{xy}=[z=xy]=\lim_{z\to 0}\frac{\sin z}{z}=1$$
and 
$$ \lim_{(x,y) \to (0,0)} \frac{\ln(1+xy)}{xy}=[z=xy]=\lim_{z\to 0}\frac{\ln(1+z)}{z}=1.$$
Is the way I calculated the limits for $(x,y)\to (0,0)$ by substituting with $z=xy$ legit? 
Also, if it is... am I allowed to substitute an expression with its limit inside a limit, as in while calculating the limit, or can I only take the limits in one last step (I'm a bit confused by this exercise in general, I have solved it with Taylor series but I'm curious to know whether this works too)?
Thank you so much in advance!
 A: Your substitution is correct - substituting $z = xy$ isn't really "doing" anything, it's just giving a name to something. It's always fine to rename things.
On the other hand, replacing an expression with its limit inside a limit is not in general permissible. To take an excellent example: consider $\lim_{x \to 0}(1 + x)^{1/x}$. If we permit ourselves to take that step, then we could observe that $1 + x \to 1$ as $x \to 0$, so this is really $\lim_{x \to 0}1^{1/x}$. Since $1^y = 1$ for every $y$, this is $\lim_{x\to 0}1$, and so the limit is just $1$. On the other hand, $\lim_{x \to 0}(1 + x)^{1/x}$ is one of the definitions of $e$, so it can't be $1$!
In this particular case, it worked - but this is because you got lucky with the functions you were considering. Basically, $\ln$ "damps" small errors near and above $1$, so when you replaced $\sin(xy)$ with the slightly-different $xy$ the difference between $\ln(1 + \sin^2(xy))$ and $\ln(1 + (xy)^2)$ was even smaller. Replacing $\ln(1 + x^2y^2)$ with $x^2y^2$ was basically okay (though only because the final answer was $0$), but technically incorrect - what you should do is use little-$o$ notation here, and write $\ln(1 + x^2y^2) = x^2y^2 + o(x^2y^2)$. If you're familiar with little-$o$ notation, you should be able to demonstrate that $\frac{o(x^2y^2)}{x^2 + y^2}$ goes to zero.
A: use that $$x^2+y^2\geq 2|xy|$$ then $$\frac{\ln(1+\sin^2(xy))}{x^2+y^2}\le \frac{\ln(1+\sin^2(xy))}{2|xy|}$$ substituting $$xy=t$$ then we have $$\frac{\ln(1+\sin^2(t))}{2t}$$ with L'Hospital we can prove that $$\lim_{t\to 0}\frac{\ln(1+\sin^2(t))}{2t}=\lim_{t\to0}\frac{\sin(2t)}{2(1+\sin^2(t))}=0$$
