Prove that $\lim_{(x,y)\to(2,0)}{\frac{xy^2}{x+y^4+3}}=0$

Question

I am trying to prove that $\lim_{(x,y)\to(2,0)}{\frac{xy^2}{x+y^4+3}}=0$ using the $\epsilon$-$\delta$ limit definition. So to prove that $\lim_{(x,y)\to(2,0)}{\frac{xy^2}{x+y^4+3}}=0$ it would be enough to show that $\forall \epsilon>0 \; \exists \delta>0 : \; [\sqrt{(x-2)^2+y^2}<\delta] \implies \Big| \frac{xy^2}{x+y^2+3}\Big| < \epsilon $.

Now I can't figure out anything reasonable to do to find what $\delta$ should be. I also tried using a rectangle instead of a sphere, but didn't get anywhere with this approach either. I've been struggling for quite a while with this and I appreciate all the help I can get.

Answer 1

You may start by taking $(x,y)$ close to $(2,0)$ (This is a rough approximation) to get control over the denominator: for example Take $|x-2|<\frac12$ and $|y|<\tfrac12$. Then $x+y^2+3>1$ and so $$\left|\frac{x y^2}{x+y^4+3}\right|\leq |x|y^2$$

Now that denominator is under control, dealing with the numerator is much easier:

Given $\varepsilon>0$, let $\delta<\min(\frac12, \frac{1}{3}\sqrt{\varepsilon})$. Then $|x|y^2<\varepsilon$

Answer 2

We have that eventually

$$\frac{xy^2}{x+y^4+3}\le \frac{(x-2)^2(x+3)+(x+3)y^2}{x+3}=(x-2)^2+y^2$$

Answer 3

If you are satisfied with a rectangle, here's an argument:

For $(x,y)\in (2-\delta, 2+\delta)\times (-\delta, \delta)$ where $\delta := \min \{\sqrt\epsilon,1\}$:

$$\begin{align}\left|\frac{xy^2}{x+y^4+3}\right|&=\frac{xy^2}{x+y^4+3} \\&<\frac{(2+\delta)\delta^2}{(2-\delta)+0+3} \\&\le\delta^2\frac{2+1}{(2-1)+3} \\&<\delta^2 \\&\le \epsilon \end{align}$$