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So, my book has this homework problem that has me pulling my hair out over. It asks us to evaluate the limit below - or rather, by trying along paths of the x and y axes, to show that it doesn't exist.

$\lim\limits_{(x,y) \to (0,0)} \frac{3}{x^2+2y^2}$

When I evaluate along $y=0$ I get

$\lim\limits_{(x,0) \to (0,0)} \frac{3}{x^2+2(0)^2}$

Which to me looks like it should just evaluate to +∞, just as along $x=0$ I get

$\lim\limits_{(0,y) \to (0,0)} \frac{3}{0^2+2y^2}$

which also looks like it should evaluate to +∞.

The flimsiest thing I can get about it not existing is that the Y term increases faster than the X term, but the book's answers section tells me that the limit doesn't exist along $x=0$.

I'm probably just forgetting something really basic like a moron, but can someone beat me over the head, please?

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  • $\begingroup$ When the limit is $\pm\infty$ we say the limit doesn't exist (in $\Bbb{R}$). $\endgroup$ Commented Feb 22, 2017 at 5:02
  • $\begingroup$ It's a matter of convention whether or not a diverging limit can be said "to exist." $\endgroup$
    – Ryan
    Commented Feb 22, 2017 at 5:25

3 Answers 3

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We have
$$ \lim\limits_{(x,y) \to (0,0)} \frac{3}{x^2+2y^2}=\infty $$ thus the limit does not exist as a real number since $\mathbb{R}=(-\infty,\infty)$.

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Given any number $B>0$, we see that

$$\frac{3}{x^2+2y^2}\ge \frac{3/2}{x^2+y^2}>B$$

whenever $\sqrt{x^2+y^2}<\sqrt{\frac{3}{2B}}$. Hence, the limit fails to exist.

That is to say, the $\frac{3}{x^2+2y^2}$ does not converge.

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  • $\begingroup$ @OlivierOloa No, I didn't mean to write that. I'll edit. And thank you Olivier! -Mark $\endgroup$
    – Mark Viola
    Commented Feb 22, 2017 at 5:26
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Let $u=x^2+2y^2$ then we have,

$$\lim_{u \to 0^+} \frac{3}{u}$$

Like you said this goes to $+\infty$. Sometimes if limits tend to $\infty$ or $-\infty$, we say they do not exist. It depends on your definition of "does not exist". Personally, I prefer to say the limit goes to $+\infty$, which isn't wrong in this case.

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