prove limit doesn't exist $\lim\limits_{(x,y)\to(0,0)} \frac{1-\cos(x^2+y^2)}{(x^2+y^2)x^2y^2}$ I need to show that limit doesn't exist:
$\lim\limits_{(x,y)\to(0,0)} \frac{1-\cos(x^2+y^2)}{(x^2+y^2)x^2y^2}$
How can I show it? 
 A: Because$$\lim_{x\to0}\frac{1-\cos(2x^2)}{(2x^2)x^2x^2}=\lim_{x\to0}\frac{1-\cos(2x^2)}{2x^6}=\infty.$$
A: Hint: Take $$x=\frac{1}{\sqrt{n}},y=\frac{1}{n^2}$$, then your limit is infinity.
A: First, not that for $xy\ne 0$, we have
$$\begin{align}
\frac{1-\cos(x^2+y^2)}{x^2y^2(x^2+y^2)}&=\frac{2\sin^2(x^2+y^2)}{x^2y^2(x^2+y^2)}\\\\
&=2\left(\frac{\sin(x^2+y^2)}{x^2+y^2}\right)^2\times \left(\frac{x^2+y^2}{x^2y^2}\right)\tag1
\end{align}$$
The first term on the right-hand side of $(1)$, $\left(\frac{\sin(x^2+y^2)}{x^2+y^2}\right)^2$, tends to $1$ as $(x,y)\to(0,0)$ However, the second term, $\frac{x^2+y^2}{x^2y^2}$ tends to $+\infty$ since
$$\frac{x^2+y^2}{x^2+y^2}\ge \frac{1}{x^2}\to \infty$$
Therefore, we see that 
$$\lim_{(x,y)\to(0,0)}\frac{1-\cos(x^2+y^2)}{x^2y^2(x^2+y^2)}=\infty$$
A: Polar coordinates:
$\dfrac{1-\cos (r^2)}{r^6 \cos^2 t \sin^2 t}= 4\dfrac{1- \cos (r^2)}{r^6 \sin^2(2t)}.$
$2t \not = kπ$, $k=0,1,2,3$.
Fix $t$, and consider
$A \dfrac{1-\cos (r^2)}{r^6}$, with $A=\dfrac{4}{\sin^2(2t)}>0$.
$A \lim_{r \rightarrow 0}\dfrac{1-\cos (r^2)}{r^6}=\infty.$
