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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.

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    $\begingroup$ "I got stuck as" has a mistake with the $r$ in the denominator. $\endgroup$ – Teepeemm Dec 13 '18 at 20:31
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Note that

$$\frac1{|x|}+\frac1{|y|}\ge \frac1{|x|}\to \infty$$

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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.

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    $\begingroup$ You are right, but a few more words about why taking "individual limits of each component" can replace $(x,y)\to (0,0)$ would be helpful in the present circumstance. $\endgroup$ – hardmath Dec 13 '18 at 14:44
  • $\begingroup$ Thank you for the advice, I'll add an edit expanding on that. $\endgroup$ – M.M. Dec 14 '18 at 10:06
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Hint:

Use the fact that if $\|(x,y)\| < \epsilon$, then $|x|<\epsilon$ and $|y|<\epsilon$

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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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