$\lim_{(x,y) \to (\frac{-k\pi}{2},\frac{k\pi}{2})} \frac{|\sin(x-y)|^a}{|x^2-y^2|}$ with $a>0, k \in \Bbb Z$ Let $a>0$, $k \in\Bbb Z$ and $k \neq 0$.
Evaluate: $\lim_{(x,y) \to (\frac{-k\pi}{2},\frac{k\pi}{2})} \frac{|\sin(x-y)|^a}{|x^2-y^2|}$
The solution indicated is $+\infty$, for every $a$, but I don't understand why.
For example, if $a=2$, moving on $(\frac{-k\pi}{2}+2t,\frac{k\pi}{2}+t)$, $t \to 0$, I obtain:
$$\frac{\sin(t)^2}{|3t^2-k\pi t|}\sim \frac{t^2}{k\pi |t|}\sim \frac{|t|}{k\pi} \to 0$$
Where am I wrong?
Thanks in advance.
 A: 
Let $a>0$, $\;k\in\Bbb Z$ and $\;k\neq 0\;.\;$
Evaluate: $\lim\limits_{(x,y)\to\left(-\frac{k\pi}{2},\frac{k\pi}{2}\right)}\dfrac{|\sin(x-y)|^a}{|x^2-y^2|}$

It results that
$\begin{align}
\dfrac{|\sin(x-y)|^a}{|x^2-y^2|}&=\dfrac{|\sin(x-y+k\pi)|^a}{|x-y+k\pi|^a}\cdot\dfrac1{|x-y|}\cdot\\
&\cdot\dfrac{|x-y+k\pi|^a}{|x+y|}\;\;,
\end{align}$
for any $\;(x,y)\in\mathbb{R}^2\land y\ne\pm x\land y\ne x+k\pi\;.$
$\lim\limits_{(x,y)\to\left(-\frac{k\pi}{2},\frac{k\pi}{2}\right)}\dfrac{|\sin(x-y+k\pi)|^a}{|x-y+k\pi|^a}=\lim\limits_{t\to0}\dfrac{|\sin t|^a}{|t|^a}=1\;,$
$\lim\limits_{(x,y)\to\left(-\frac{k\pi}{2},\frac{k\pi}{2}\right)}\dfrac1{|x-y|}=\dfrac1{|k|\pi}\;,$
but
$\lim\limits_{(x,y)\to\left(-\frac{k\pi}{2},\frac{k\pi}{2}\right)}\dfrac{|x-y+k\pi|^a}{|x+y|}\;$ does not exist, indeed,
by letting $\;X=x-y+k\pi\;$ and $\;Y=x+y\;,\;$ we get that
$\lim\limits_{(x,y)\to\left(-\frac{k\pi}{2},\frac{k\pi}{2}\right)}\dfrac{|x-y+k\pi|^a}{|x+y|}=\lim\limits_{(X,Y)\to(0,0)}\dfrac{|X|^a}{|Y|}\;.$
If $\;\lim\limits_{(X,Y)\to(0,0)}\dfrac{|X|^a}{|Y|}\;$ existed, then the value of the limit would be the same on every subset $\;S\;$ of the domain of the function $\;\dfrac{|X|^a}{|Y|}\;$ for which $\;(0,0)\;$ is a limit point of $\;S\;,\;$ but
$\;\lim\limits_{(X,Y)\to(0,0)\\Y=m|X|^a}\dfrac{|X|^a}{|Y|}=\dfrac1{|m|}\;$
is not the same in fact it depends on $\;m\;,\;$ hence
$\lim\limits_{(x,y)\to\left(-\frac{k\pi}{2},\frac{k\pi}{2}\right)}\dfrac{|x-y+k\pi|^a}{|x+y|}=\lim\limits_{(X,Y)\to(0,0)}\dfrac{|X|^a}{|Y|}\;$
does not exist, consequently
even $\;\lim\limits_{(x,y)\to\left(-\frac{k\pi}{2},\frac{k\pi}{2}\right)}\dfrac{|\sin(x-y)|^a}{|x^2-y^2|}\;$ does not exist.
