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:
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.
$$\begin{align} L&=\lim_{(x,y)\to(a,a)}{\frac{(x-y)a+(a-x)y-(a-y)x}{(x-y)(a-x)(a-y)}} \\[1em] &=\lim_{(x,y)\to(a,a)}{\frac{0}{(x-y)(a-x)(a-y)}} \\[1em] \end{align}$$
which is (again) indeterminate if $(a,a)$ is approached along the line $y=x$.
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
$$\binom{n}{2}a^{n-2}.$$
