Does the limit $\lim_{(x,y)\to(0,0)}\frac{x^2y}{x^2+y^4}$ exist?

Question

I am trying to evaluate $$\lim_{(x,y)\to(0,0)}\frac{x^2y}{x^2+y^4}$$

I was thinking of using $$0\leq\frac{x^2y}{x^2+y^4}<\frac{(x^2+y^4)\cdot y}{x^2+y^4}=y$$ which tends to as $(x,y)\to(0,0)$, which means that the limit is $0$ by the squeeze theorem. Is that correct?

that doesn't sound right... for small values of $y$, $(x^2+y^4)y$ is smaller than $x^2y$ — Saketh Malyala, Jun 10 '19 at 22:01
Does this answer your question? Multivariable limit proof: $\lim_{(x,y)\rightarrow (0,0)}\frac{\left|x\right|^a\left|y\right|^b}{\left|x\right|^c + \left|y\right|^d} = 0$ — Arnaud D., Oct 20 '20 at 11:42

score 5 · Accepted Answer · answered Jun 10 '19 at 22:04

5

$$\left|\frac{x^2y}{x^2+y^4}\right|\le\frac{x^2|y|}{x^2}=|y|\xrightarrow[(x.y)\to(0,0)]{}0$$

answered Jun 10 '19 at 22:04

DonAntonio

211,718
17
136
287

Does the limit $\lim_{(x,y)\to(0,0)}\frac{x^2y}{x^2+y^4}$ exist?

1 Answers1

Linked