How to prove that $\lim_{(x, y) \to (0, 0)} \left(\frac{1}{|x|} + \frac{1}{|y|}\right) = 0$?

Question

Prove that $$\lim_{(x, y) \to (0, 0)} \left(\frac{1}{|x|} + \frac{1}{|y|}\right) = 0$$
I couldn't prove this. Please suggest a solution.

Answer 1

Nobody can suggest a solution, because the statement is false.

If you approach $(0,0)$ with the sequence $(1/n,1/n)$ the limit is $\infty$. Therefore the given limit is certainly not $0$.

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If your assignment was proving that $$ \lim_{(x,y)\to\infty}\left(\frac{1}{|x|}+\frac{1}{|y|}\right)=0 $$ you need to show that, for every $\varepsilon>0$ there exists $M$ such that, for $\sqrt{x^2+y^2}>M$ (and $xy\ne0$) we have $$ \frac{1}{|x|}+\frac{1}{|y|}<\varepsilon $$ Now, for $a>0$ and $b>0$, we have $$ \sqrt{ab}\le\frac{a+b}{2} $$ so $$ \frac{a+b}{ab}\le\frac{4}{a+b}\le\frac{8}{\sqrt{a^2+b^2}} $$ (the second inequality is readily verified). Thus, taking $M=8/\varepsilon$, for $\sqrt{x^2+y^2}>M$ we have $$ \frac{8}{\sqrt{x^2+y^2}}<\frac{8}{M}=\varepsilon $$ and we are done: $$ \varepsilon=\frac{8}{M}>\frac{8}{\sqrt{x^2+y^2}} \ge\frac{4}{|x|+|y|}\ge\frac{|x|+|y|}{|x|\,|y|} =\frac{1}{|x|}+\frac{1}{|y|} $$