Change of variables from multiple to single Consider the following limit calculation:
$$ \lim_{(x,y)\to (0,0)} \frac{\sin(x^2+y^2)}{x^2+y^2} = \lim_{t\to 0} \frac{\sin t}{t} = 1$$
How can one justify this change of variables from multiple to single? When can we do it (and when can't we?)
Thanks.
 A: The function was rotationally symmetric and only depended on $t=x^2+y^2$ so $x^2+y^2$ could have been replaced by $t$.
Why it can be done may be seen by looking at the meaning of the "limit", especially the two-dimensional one. The limit $A\to B$ (one-dimensional or two-dimensional or higher-dimensional) exists and is equal to $L$ if for any, arbitrarily small $\epsilon\gt 0 $ ("desired accuracy"), one can find a correspondingly small $\delta \gt 0$ such that in the whole vicinity $A\in B\pm \delta$, i.e. the $\delta$-radius vicinity of $B$, the value of the function belongs to the interval $L\pm \epsilon$.
Now, in your case, the replacement is allowed because the $\delta$-radius vicinity of the point $(0,0)$ in the two dimensional case (the disk of radius $\delta$) is exactly the set mapped to the (positive side) of the vicinity of $0$ for $t$ of the radius $\delta^2$. The relevant neighborhoods get mapped to each other perfectly.
More precisely, you should have written the limit for $t$ as the limit $t\to 0^+$ which only requires the existence of the limit for positive small values of $t$ – because $t=x^2+y^2$ is guaranteed not to be negative. But in your case it doesn't matter because the function of $t$ is even, so the limit from one side is the same as the overall limit around zero.
A: By using polar substitution.
That is,
Let $x=r\cos A$ and $y=r\sin A$, implying $r^2 = x^2 + y^2$. Then $(x,y)\to(0,0)\implies r\to0$. Note! the existence or otherwise of the limit should be independent of the value of $A$.
A: If the limit exists, then you can choose any path (parametristation of x = $\frac{\sqrt{2}}{2} t$ and y = $\frac{\sqrt{2}}{2} t$ in this case) to the limit point $(0,0)$ you like, and you'll end up with exactly the same result.
This is kind of moot point though since the limit exists if and only if every continuous path taken to the limit point results in the same value for the limit.
In general we say that $lim_{\textbf{x} \rightarrow \textbf{a} } f(\textbf{x}) $ exist and equals L if:
$\forall \epsilon > 0$ $\exists \delta > 0$ in such a way that $||\textbf{x} - \textbf{a} || < \delta \rightarrow ||f(\textbf{x}) - L|| < \epsilon$.
Or in human speak: the limit exists if, when we get close to the limit point, the function values get close to the limit value as well.
