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.   
 A: 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 $y=x$.
A: Note that this function is not defined when $x=y$ or $x=a$ or $y=a.$ In other words, it's not defined on the union of those three lines, all of which pass through $(a,a).$
Let $x=a+h, y = a +k.$ We assume both $h,k\ne 0$ and $h\ne k$ in view of the comments above. Then the expression equals
$$\frac{(h-k)a^n -h(a+k)^n +k(a+h)^n}{(h-k)hk}.$$
We want the limit of this as $(h,k) \to (0,0),$  subject to the restrictions mentioned.
It's easy to see the numerator vanishes when $n=1,$ so the limit is $0$ in that case. For $n>1,$ expand those $n$th powers using the binomial theorem. There's some nice cancellation. We get
$$\frac{\sum_{m=2}^{n}\binom{n}{m}a^{n-m}hk(h^{m-1}-k^{m-1})}{(h-k)hk}.$$
Now $h-k$ divides $h^{m-1}-k^{m-1}$ for $m=2,\dots, n.$ Thus the denominator is a factor in each summand. As $(h,k)\to (0,0),$ the only summand that contributes to the limit is the $m=2$ term. (The other summands look like $(h-k)hk$ times positive powers of $h,k.$) The desired limit is thus
$$\binom{n}{2}a^{n-2}.$$
