# 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.

• "I got stuck as" has a mistake with the $r$ in the denominator. – 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.

$$\lim_{(x,y)\to(0,0)}{\bigg(\frac{1}{|x|}+\frac{1}{|y|}\bigg)}=\lim_{(x,y)\to(0,0)}{\bigg(\frac{1}{|x|}\bigg)}+\lim_{(x,y)\to(0,0)}{\bigg(\frac{1}{|y|}\bigg)}$$

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.

• 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. – hardmath Dec 13 '18 at 14:44
• Thank you for the advice, I'll add an edit expanding on that. – M.M. Dec 14 '18 at 10:06

Hint:

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

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

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

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

$$\dfrac{1}{|x|} +\dfrac{1}{|y|}.$$