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How to compute the following limit: $$\lim_{(x,y)\to (0,0)}\frac{1+x-y}{x^2+y^2}$$ ?

The teacher's answers is "the limit doesn't exist". But when replace the variables, the value is $$\frac{1 + 0 - 0}{0^2+0^2}=\frac{1}{0}=\infty.$$ Which one is correct?

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  • $\begingroup$ $\infty$ is usually taken to be a case where the limit does not exist. $\endgroup$ – Cameron Williams May 27 '17 at 18:39
  • $\begingroup$ $\infty$ ins't strictly a limit $\endgroup$ – Hagen von Eitzen May 27 '17 at 18:39
  • $\begingroup$ Can I say that the limit does not exist, right? Thank you guys! $\endgroup$ – ComplexityAlg May 27 '17 at 18:43
  • $\begingroup$ When $x=y \ne 0$ the function is $1/(x^2+y^2)=1/|x|^2.$ So if we approach $(0,0)$ along the line $x=y$, the function $\to \infty.$ $\endgroup$ – DanielWainfleet May 27 '17 at 23:41
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The limit is $\infty$ for $(x,y)$ close to $(0,0)$ you have that $1+x-y\ge\frac12$ and so $$\frac{1+x-y}{x^2+y^2}\ge \frac{1}{2(x^2+y^2)}\ge M$$ provided $0<x^2+y^2\le \frac1{2M}$.

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  • $\begingroup$ How can I find this 1/2? Anyway, thank you guy! $\endgroup$ – ComplexityAlg May 27 '17 at 18:45
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    $\begingroup$ You have to use the definition of limit. You have $|x|+|y|\le 2\sqrt{x^2+y^2}\le \delta$, so you can take $\delta<\frac14$ or anything you want. $\endgroup$ – Gio67 May 27 '17 at 19:10
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If $|x|<1/3$ and $|y|<1/3$ then $$\frac{1+x-y}{x^2+y^2}\geq \frac{1-|x|-|y|}{x^2+y^2}\geq \frac{1/3}{x^2+y^2}\to+\infty$$ as $(x,y)\to (0,0)$. Therefore the limit is $+\infty$.

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  • $\begingroup$ How can I find this 1/3? Anyway, thank you guy! $\endgroup$ – ComplexityAlg May 27 '17 at 18:45
  • $\begingroup$ @ComplexityAlg You can replace $1/3$ with any real number $r$ such that $0<r<1/2$ (so $1-r-r>0$). $\endgroup$ – Robert Z May 27 '17 at 19:05
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plugging $$x=1/n,y=1/n$$ in the term $$\frac{1+x-y}{x^2+y^2}$$ we get $$\frac{1}{2}{n^2}$$ and this goes to infinity,therefore the Limit doesn't exist.

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  • $\begingroup$ What does "1/n" mean? Thank you guy! $\endgroup$ – ComplexityAlg May 27 '17 at 18:46
  • $\begingroup$ the term $$\frac{1}{n}$$ tends to Zero if $n$ tends to infinity $\endgroup$ – Dr. Sonnhard Graubner May 27 '17 at 18:48
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I would say that $$\frac 1 0$$ gives an $NaN$, because if we consider $$\frac x y$$ to be an operation where we count the number of times we need to subtract $y$ from $x$ to get $0$, if $y=0$ then even after an infinite number of subtractions we still have not reached $0$. Now, I would say that the limit $$\lim_{y\to0} \frac x y$$ has a result equal to $\infty$ rather than $NaN$ because we can imagine that $y$ is not quite $0$ but is infinitely close, so after an infinite number of subtractions we do eventually come to reach $0$.

Cheers!

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