<s>[INST] 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?
 [/INST] 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. 
</s>