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.
 A: Note that
$$\frac1{|x|}+\frac1{|y|}\ge \frac1{|x|}\to \infty$$
A: Take the individual limits of each component. x and y are tending towards zero (the direction doesn't matter due to the magnitudes) so $\frac{1}{|x|}$ and $\frac{1}{|y|}$ both tend to infinity, and as a result so does their sum.
EDIT:
As suggested in the comments, this is an expansion as to why you can seperate the limits.
$$\lim_{(x,y)\to(0,0)}{\bigg(\frac{1}{|x|}+\frac{1}{|y|}\bigg)}=\lim_{(x,y)\to(0,0)}{\bigg(\frac{1}{|x|}\bigg)}+\lim_{(x,y)\to(0,0)}{\bigg(\frac{1}{|y|}\bigg)}$$
Examining the LHS we see that the first limit is independent of y, and the second independent of x. As a result we can rewrite the equation as follows:
$$=\lim_{x\to0}{\bigg(\frac{1}{|x|}\bigg)}+\lim_{y\to0}{\bigg(\frac{1}{|y|}\bigg)}$$
As stated above, the magnitudes mean that both limits tend to infinity as x and y tend to zero from either direction, so their sum is necessarily also infinity.
A: Hint:
Use the fact that if $\|(x,y)\| < \epsilon$, then $|x|<\epsilon$ and $|y|<\epsilon$
A: 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 $
$\dfrac{1}{|x|} +\dfrac{1}{|y|}.$
