<s>[INST] Find existence of limit Show that $\left(\frac{1}{|x|} +\frac{1}{|y |}\right)$ tends to infinity as $(x,y)$ tends to $(0,0)$.
I have used $x=r\cos \alpha$ and $y=r\sin \alpha$,  $r>0$.But I got stuck as $\left(\frac{1}{|\cos \alpha|} +\frac{1}{ |r\sin \alpha|}\right) \left(\frac{1}{r}\right)$  lies between $\frac{1}{r}$ and infinity.
 [/INST] We have:
$(x,y) \rightarrow 0.$
Let $\epsilon_n =1/n$, $n$ positive integer, be given.
Then $|(x^2+y^2)^{1/2}| < \delta_n$ $(=\epsilon_n)$ implies 
$|(x^2+y^2)^{1/2}| < \epsilon_n =1/n$, or
$n=1/\epsilon_n \lt \dfrac{1}{(x^2+y^2)^{1/2}}.$
Note:
$(x^2+y^2)^{1/2} \ge |x|$, and 
$(x^2+y^2)^{1/2} \ge |y|.$
Then
$\dfrac{2}{(x^2+y^2)^{1/2}} \le \dfrac{1}{|x|}+\dfrac{1}{|y|}.$
And finally :
$|(x^2+y^2)^{1/2}| \lt \delta_n$ implies
$2n =2/\epsilon_n \lt \dfrac{2}{(x^2+y^2)^{1/2}} \le $
$\dfrac{1}{|x|} +\dfrac{1}{|y|}.$
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