limit of the function of two variables ?? Here is the function :
\begin{equation}
f(x,y)=a^2 \left(\frac{x}{a^2-3 x^2}-\frac{y}{3 y^2-a^2}\right).
\end{equation}
We know that $y\geq x \geq 0$, $~a$ is a constant and $a^2=x^2+xy+y^2$.
I want to derive the limit of the function in the case $x\rightarrow a/\sqrt{3}=r_0$ and  $y\rightarrow a/\sqrt{3}=r_0$.
One method is to take $x=r_0-\delta$ and $y=r_0+\delta$ and $a=\sqrt{3}r_0$. Then substituting them into $f(x,y)$, I can obtain 
\begin{equation}
f(r_0,\delta)=-\frac{2 r_0^3}{4 r_0^2-\delta ^2}.
\end{equation}
Thus in the limit of $\delta \rightarrow 0$, we can get the result $-r_0/2$.
The another method is to substitute $a^2=x^2+xy+y^2$ into $f(x,y)$ directly, I get 
\begin{equation}
-\frac{(x+y) \left(x^2+x y+y^2\right)}{(2 x+y) (x+2 y)},
\end{equation}
then set $x=y=r_0$. I get the limit is $-2r_0/3$.
Obviously, the two results are different. So which method is reliable?
 A: Write $x:= ua$, $y:=v a$. We then have to investigate
$$\hat f(u,v)=a\left({u\over 1-3u^2}+{v\over 1-3v^2}\right)$$
under the constraint
$$u^2+uv+v^2=1\ .$$
The constraint defines an ellipse in the $(u,v)$-plane whose axes are at $45^\circ$ with respect to the coordinate axes. A parametrization of this ellipse is given by
$$u(\phi)={1\over\sqrt{3}}\cos\phi-\sin\phi,\quad v(\phi)={1\over\sqrt{3}}\cos\phi+\sin\phi\qquad(-\pi\leq\phi\leq\pi)\ .$$
The function $g(\phi):=\hat f\bigl(u(\phi),v(\phi)\bigr)$ simplifies to
$$g(\phi)=-{2a\over\sqrt{3}}\ {\cos\phi\over 1+2\cos(2\phi)}\ .$$
The point $x=y={a\over\sqrt{3}}$ in question corresponds to the point $u=v={1\over\sqrt{3}}$ on said ellipse, hence to $\phi=0$ on the parameter axis. This means that we have to compute
$$\lim_{\phi\to0} g(\phi)=-{2a\over 3\sqrt{3}}\ .$$
A: The first method makes no sense: once you decide that $x=r_0-\delta$, $y=r_0+\delta$, you have
$$
x^2+xy+y^2=(r_0+\delta)^2+(r_0-\delta)(r_0+\delta)+(r_0-\delta)^2=3r_0^2+\delta^2=a^2+\delta 
$$
and your condition $x^2+xy+y^2=a^2$ is not satisfied. 
In the second method you are not showing your work so I cannot comment on it, but you are getting the right result. 
