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.

  • 3
    $\begingroup$ "I got stuck as" has a mistake with the $r$ in the denominator. $\endgroup$ – Teepeemm Dec 13 '18 at 20:31

Note that

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


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.


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:


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


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


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

$n=1/\epsilon_n \lt \dfrac{1}{(x^2+y^2)^{1/2}}.$


$(x^2+y^2)^{1/2} \ge |x|$, and $(x^2+y^2)^{1/2} \ge |y|.$


$\dfrac{2}{(x^2+y^2)^{1/2}} \le \dfrac{1}{|x|}+\dfrac{1}{|y|}.$

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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