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Show of the limit exists or does not $$\lim_{(x,y,)\to (0,0)} \frac {x \sqrt{3x^2+7y^2}}{|y|}$$

{EDIT I first used "two-path" technique but couldn't figure out any such paths so that the limit would give different values.

Then I put it on Wolfram Alpha and it showed that the limit is 0.}

I couldn't make any useful inequality to use squeeze theorem. Then I tried polar coordinates and got : $$ \frac{r\cos \theta \sqrt{3\cos^2 \theta + 7 \sin^2 \theta}}{|\sin \theta|}$$

But the $|\sin \theta|$ in the denominator is stopping me there to put $r = 0$ and evaluate the limit.

How can I solve it then?

Edit3 This just shows that the limit can't be $0$, not that the limit doesn't exist. Still I'm letting it be as I thought it'd be relevant[Edit 2

Using @Kavi's hint: If the limit were to exist and be equal to $0$, for $\epsilon = 1$ $\exists \delta \gt 0$ such that $$\frac {|x| \sqrt{3x^2+7y^2}}{|y|} \lt 1 \\ \implies |x| \sqrt{3x^2+7y^2} \lt |y|$$ for all $(x,y)$ such that $\sqrt{x^2+y^2} \lt \delta$.

Now from above $y\to 0$ implies $$ |x| \sqrt{3x^2} \le 0$$ for all $|x| \lt \delta$. But if we take $x\in (0,\delta)$ then $|x| \sqrt{3x^2} \gt 0$ which is a contradiction. So the limit doesn't exist.

Is this argument right?]

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    $\begingroup$ It is right and that is exactly what I wrote in my answer. $\endgroup$ Apr 23 at 12:20
  • $\begingroup$ @KaviRamaMurthy actually what you commented about using $y=x^2$ is the only right approach. What I've done in the edit shows that the limit is not 0. It doesn't show that the limit doesn't exist. So I think I've done it wrong. $\endgroup$
    – Itachi
    Apr 23 at 12:29
  • $\begingroup$ Are you happy with my answer now? $\endgroup$ Apr 23 at 12:32
  • $\begingroup$ Yes I am happy with your answer now. $\endgroup$
    – Itachi
    Apr 23 at 12:43

3 Answers 3

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The claim supposedly made by Wolfram alpha is false! First objection is that the function is not defined on the $x-$ axis. Eevn if you avoid the $x-$ axis the result is false. If it is true then there exists $\delta >0$ such that $|x|\sqrt {3x^{2}+y^{2}} <|y|$ whenever $\|(x,y)\|<\delta$. Let $y \to 0$ to get $|x|\sqrt {3x^{2}} \leq 0$ whenever $|x|<\delta$. Of course, this is false.

To show that the limit does not exist consider the limit along $y=x^{2}, x>0$ and $y=x^{2}, x<0$.

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  • $\begingroup$ Okay the orginal question was to show if the limit exists. I put it into wolfram alpha but it showed to be 0. I've used "two-path" methods but can't figure out that limit doesn't exist. Can you give me an example of two paths that lead the limit to two different values? $\endgroup$
    – Itachi
    Apr 23 at 11:43
  • $\begingroup$ You must have done something wrong in entering data on Wolfram alpha. Check again. Anyway since the function is not even defined in any neighborhood of $(0,0)$ the statement is false and you don't have to produce paths along which the limits are different. If it is supposed to be limit as $(x,y) \to (0,1)$ then the statement is correct @Itachi $\endgroup$ Apr 23 at 11:55
  • $\begingroup$ no I've checked multiple times but I have put the date correctly. But yes you're right that the function isn't defined on $x$-axis. So as you said is it enough to write this to show that the limit doesn't exist? $\endgroup$
    – Itachi
    Apr 23 at 12:01
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    $\begingroup$ @Itachi The limit along $y=x^{2}$ through positive $x$ is $\sqrt 3 $ and the limit along the same path through negative values of $x$ is $-\sqrt 3$. $\endgroup$ Apr 23 at 12:05
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    $\begingroup$ @KaviRamaMurthy The function $f$ is defined by $f:\mathbb{R}\times\mathbb{R}^{*}\to \mathbb{R}$ with $(x,y)\mapsto \frac{x\sqrt{3x^{2}+7y^{2}}}{|y|}$ so with the natural domain $f$ is well-defined.Why then do we care about the $X-$ axis? And Indeed, Wolfram give $0$. $\endgroup$ Apr 23 at 12:13
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It's not necessary to use two paths! On the x-axis, y= 0, the fraction is $\frac{X\sqrt{3x^2}}{0}$ which does not exist for any x!

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  • $\begingroup$ But the path $r(t)=(t,0)$ is not in the domain of the function, so why are you considering that path? $\endgroup$ Apr 23 at 12:19
  • $\begingroup$ To show that a limit doesn't exists you need to consider points in the domain of the function. $\endgroup$
    – jjagmath
    Apr 23 at 12:24
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A necessary condition for the limit $\lim_{(x,y)\to (a,b)}f(x,y)$ to exist is that there exists a neighborhood of the point $(a,b)$ on which the function is well-defined; that is there exists $\delta>0$ such that $f(x,y)$ exists for every $(x,y)$ such that $|x-a|+|y-b|< \delta$.

In the given problem, every neighborhood of the origin contains the points $(x,0)$ on which the function is not well-defined.

If you prefer checking the limit on a path passing through the origin take $y=x^4$ on which the limit does not exist. And we are done because you could also check the limit on the path $x=0$ and find its values thereon is $0$.

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