Prove that $ \displaystyle \lim_{(x,y) \to (0,0)} \frac{x^2y^2}{x^3 + y^6} = 0 $

Question

I am trying to show that $$ \displaystyle \lim_{(x,y) \to (0,0)} \frac{x^2y^2}{x^3 + y^6} = 0 $$ Unfortunately, I am stuck in the math. I tried squeeze theorem, and $\varepsilon$-$\delta$ but unsuccessfully.

Any suggestions on approaches?

Try the limits along the paths $t\mapsto (kt^2,t)$ with $k\in \mathbb R$. — Git Gud, Feb 20 '16 at 23:32

Martín-Blas Pérez Pinilla · Answer 1 · 2016-02-21T08:48:38.050

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Is false: the quotient isn't defined in the subset $x^3+y^6 = 0$, i.e., $x = -y^2$.

EDIT: Wolfram Alpha does the limit definitely wrong. An even excluding the bad subset, $$\lim_{n\to\infty}f(-1/n^2,1/n+1/n^2) = \infty.$$

edited Feb 21 '16 at 08:48

answered Feb 20 '16 at 23:22

Martín-Blas Pérez Pinilla

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This is peculiar...WolframAlpha tells me this evaluates to 0. What might be the issue? – Orest Xherija Feb 21 '16 at 06:42
@OrestXherija, I will check later. – Martín-Blas Pérez Pinilla Feb 21 '16 at 07:54

Prove that $ \displaystyle \lim_{(x,y) \to (0,0)} \frac{x^2y^2}{x^3 + y^6} = 0 $

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