Finding the next limit How to calculate $\displaystyle\lim_{(x,y)\rightarrow(a,a)}\frac{(x-y)a^n+(a-x)y^n-(a-y)x^n}{(x-y)(a-x)(a-y)}$ where $a\neq 0?$ 
I have the following idea:the factors on the denominator are roots of the expression on numerator because nullify it. So, the polinomyal has these three roots and then, will have a polinomyal expression who has no problem to calculate the limit. Is this idea works? 
But if this is true, how can I compute the limit value? An idea to solve this easier?
Thanks in advance.   

In order for the limit to exist it must have the same value along all paths that approach the point $(a,a)$. We will find two paths that disagree for the limit in question.

So first consider the limit along the parametric curve (a line) $x=a-h,y=a+h$, and thus let $h\to0$. We have:


*

*$x-y=-2h$

*$a-y=-h$

*$a-x=h$


So then assuming $n\ge4$
$$\begin{align}
L_1&=\lim_{h\to0}\frac{-2ha^n+h(a+h)^n+h(a-h)^n}{(-2h)(h)(-h)} \\[1em]
   &=\lim_{h\to0}\frac{[(a+h)^n-a^n]+[(a-h)^n-a^n]}{2h^2} \\[1em]
\end{align}$$
By the binomial theorem, the numerator is:
$$\begin{align}
N&=[(a^n+na^{n-1}h+\binom{n}{2}a^{n-2}h^2+\binom{n}{3}a^{n-3}h^3)^n+O(h^4)-a^n]\\&+[(a^n-na^{n-1}h+\binom{n}{2}a^{n-2}h^2-\binom{n}{3}a^{n-3}h^3)^n+O(h^4)-a^n] \\
&=2\binom{n}{2}a^{n-2}h^2+O(h^4)
\end{align}$$ 
So 
$$\begin{align}
L_1&=\lim_{h\to0}{\frac{2\binom{n}{2}a^{n-2}h^2+O(h^4)}{2h^2}} \\[1em]
   &=\lim_{h\to0}\left\{\binom{n}{2}a^{n-2}+O(h^2)\right\} \\[1em]
   &=\binom{n}{2}a^{n-2} \\[1em]
\end{align}$$
This also holds for $2\le n<4$ as higher order terms fall away to zero. 

Now consider the limit along the line $y=x$. Then
$$\require{cancel}\begin{align}
L_2&=\lim_{x=y\to a}\frac{(x-y)a^n+(a-x)y^n-(a-y)x^n}{(x-y)(a-x)(a-y)} \\[1em]
   &=\lim_{x=y\to a}\frac{(x-y)a^n+(x-a)(x^n-y^n)}{(x-y)(a-x)(a-y)} \\ \end{align}$$
For $n\ge2$, the factorisation $x^n-y^n=(x-y)(x^{n-1}+x^{n-2}y+\cdots+xy^{n-2}+y^{n-1})$
means that there is a factor of $(x-y)$ in the numerator and the denominator, so
$$\begin{align}
L_2&=\lim_{x=y\to a}\frac{a^n+(x-a)(x^{n-1}+x^{n-2}y+\cdots+xy^{n-2}+y^{n-1})}{(a-x)^2} \\[1em] 
&=\lim_{x\to a}{\frac{a^n}{(a-x)^2}}\qquad(a^n\text{ dominates the second term in the numerator}) \\[1em]
&=\begin{cases}+\infty,&\text{if }a>0 \\-\infty,&\text{if }a<0 \end{cases}
\end{align}$$

The fact that $L_1\ne L_2$ is enough to show that the original limit does not exist, not even in the sense that it is infinite.

Addendum: $n=1$
$$\begin{align}
L&=\lim_{(x,y)\to(a,a)}{\frac{(x-y)a+(a-x)y-(a-y)x}{(x-y)(a-x)(a-y)}} \\[1em]
 &=\lim_{(x,y)\to(a,a)}{\frac{0}{(x-y)(a-x)(a-y)}} \\[1em]
\end{align}$$
which is (again) indeterminate if $(a,a)$ is approached along the line $y=x$.
