Can someone please prove this limit via the squeeze theorem Can someone please prove that this limit exists using squeeze theorem?
$$\lim_{x,y\to 0,0}\frac{5x^2y}{x^2+8y^2}.$$
Another question I've to ask is for $$y = x^2$$ can we not prove that the limit does not exist? If the case is true then the limit becomes: $$\lim_{x\to 0}\frac{5x^4}{x^2+8x^4}.$$.
Can't that limit be solved using L'Hospital's rule and get a value that is not 0?
(Just for reference when we approach from both the axis, the limit is 0).
I'm sorry if my framing of the question is messy but
TLDR: I saw this question somewhere which shows that a limit exists but when I tried to use different methods of approaching the limit it gave me different answers.
 A: You have for $(x,y) \neq (0,0)$
$$0 \le \left\vert \frac{5x^2y}{x^2+8y^2}\right\vert \le \left\vert\frac{5x^2y}{x^2+y^2}  \right\vert\le \frac{5}{2} \vert x \vert \left\vert\frac{\vert x y \vert}{x^2+y^2}  \right\vert \le \frac{5\vert x \vert }{2}$$
as $\vert x y \vert \le \frac{x^2+y^2}{2}$ for all $(x,y) \in \mathbb R$.
As $\lim\limits_{(x,y) \to (0,0)} \vert x \vert = 0$, you get the desired conclusion by the squeeze theorem.
A: $$\left|\frac{5x^2y}{x^2+8y^2}\right|=5|y|\left|\frac{x^2}{x^2+8y^2}\right|\le 5|y|\to0.$$
A: We have
$$\left|\frac{5x^2y}{x^2+8y^2}\right|=\frac{5x^2|y|}{x^2+8y^2}\le \frac{5x^2|y|+5|y|y^2}{x^2+y^2}= 5|y|\frac{x^2+y^2}{x^2+y^2}=5|y| \to 0$$
or also more simply
$$\left|\frac{5x^2y}{x^2+8y^2}\right|
=5|y|\frac{x^2}{x^2+8y^2}
\le5|y| \to 0$$
A: 0)$(x,y)\not =(0,0)$;
1)$x=0;$ $y\not=0;$ The limit $=0$;
2)$y=0$; $x\not=0$; The limit $=0$;
3)$x,y \not =0$;
Then
$|\frac{5x^2y}{x^2+8y^2}| \lt \frac{5x^2|y|}{x^2} =5|y| \rightarrow 0.$
A: Hint:
Use polar coordinates:
$$\lim_{x,y\to 0,0}\frac{5x^2y}{x^2+8y^2}=\lim_{r\to 0}\frac{r^2\cos^2\theta\cdot r\sin\theta}{r^2\cos^2\theta+8r^2\sin^2\theta},$$
and observe that $\;|\cos^2\theta\sin\theta|\le 1$, whereas $\:\cos^2\theta+8\sin^2\theta=1+7\sin^2\theta\ge 1$.
