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

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?]

• It is right and that is exactly what I wrote in my answer. Apr 23, 2022 at 12:20
• @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. Apr 23, 2022 at 12:29
• Are you happy with my answer now? Apr 23, 2022 at 12:32
• Yes I am happy with your answer now. Apr 23, 2022 at 12:43

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 $$y=x^{2}, x<0$$.

• 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? Apr 23, 2022 at 11:43
• 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 Apr 23, 2022 at 11:55
• 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? Apr 23, 2022 at 12:01
• @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 $-\sqrt 3$. Apr 23, 2022 at 12:05
• @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 $0$. Apr 23, 2022 at 12:13

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!

• But the path $r(t)=(t,0)$ is not in the domain of the function, so why are you considering that path? Apr 23, 2022 at 12:19
• To show that a limit doesn't exists you need to consider points in the domain of the function. Apr 23, 2022 at 12:24

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 $$0$$.