Arguing the limit of a function using epsilon delta

Question

So I have some function f(x,y) where

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

and I have used other ways to determine that $\lim \limits_{x,y \to 0,0} {f(x,y)}$ might be $0$. So now I want to use the epsilon-delta definition to prove that the limit is $0$.

What I have so far is that $\delta > \sqrt{x^2 + y^2} > 0$ and $\epsilon > \frac{3|x|y^2}{x^2+y^4}$. My argument is that because $\sqrt{x^2+y^2}$ is always positive for $(x,y)$ in the domain of $f(x,y)$ then there exists a $\delta > 0$ for all $\epsilon > 0$ and therefore the limit is $0$.

I was wondering if there is a flaw in this logic and if there is how it should be corrected. Thank you.

Answer 1

I thought it might be instructive to present an approach that uses a well-known inequality to facilitate a $\delta-\epsilon$ proof that demonstrates that the limit fails to exist. To that end, we proceed.

Note from the AM-GM inequality, we have $$x^2+y^4\ge 2\sqrt{x^2y^4}=2|x|y^2$$with equality when $x=y^2$.

Hence, the best we can hope for is that

$$\frac{3xy^2}{x^2+y^4}\le \frac32$$

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Now, if $x=0$ (or $y=0$), then $\lim_{y\to 0}\frac{3(0)y^2}{(0)^2+y^4}=0$.

We proceed to show that the limit

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

and hence the limit does not exist.

First, take $\epsilon=1$. Then, for all $\delta>0$, we take $x^2=y^4$ such that $0<\sqrt{x^2+y^2}<\delta$ and

$$\left|\frac{3xy^2}{x^2+y^4}\right|=\frac32>\epsilon$$

Hence, the limit is not $0$.

Inasmuch as the value of the limit depends on the manner in which it is taken, the limit fails to exist.

Answer 2

If you head towards the origin along the curve $x = y^2$ then your limit looks like:

$$\frac{3xy^2}{x^2 + y^4} = \frac{3y^4}{2y^4} = \frac{3}{2}$$

which converges to $3/2$. Compare this with the limit approaching the origin along $x= y$ to derive a contradiction; thus, the limit does not exist.