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?]
Best Answer
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 $y=x^{2}, x<0$.