I'm struggling to find if the following limit exists:
$\lim_{{(x,y) \to (0,0)}} \frac{x^2y^2}{x^2+y^4}$
Intuitively, I believe it does not, but every direction I approach it from ($y=x$, $y=x^2$,$y=0$, $x=0$ etc.) seems to provide the same limit of zero. Any help is very much appreciated, as well as any tips on finding a suitable direction to prove a limit doesn't exist.(Excluding the polar coordinate conversion method)
Notice that $$\frac{x^2y^2}{x^2+y^4} \leqslant \frac{x^2y^2}{x^2}=y^2$$ and hence the function is bounded near $(0,0)$ by $y^2$.
So, the limit is $0$.
There is an alternative but less evident method $$|xy^2|\le {1\over 2}(x^2+y^4)$$ Hence $${x^2y^2\over x^2+y^4}={|x|y^2\over x^2+y^4}|x|\le {1\over 2}|x|$$ Thus the limit is $0$ if $x\to 0.$ In other solutions the limit is $0$ if $y\to 0.$
If in neighborhood of some point $a$ you have that $|f(x) - L| \leq g(x)$ and $\displaystyle \lim_{x\to 0} g(x) = 0$, you have that $\displaystyle \lim_{x\to a} f(x) =L$. In this case, as it was mentioned in the comments, we have that $$ \left| f(x,y) -0 \right|=\frac{x^2y^2}{x^2+y^4} \leq \frac{(x^2+y^4)y^2}{x^2+y^4} = y^2 \to 0, \quad \text{as } (x,y)\to (0,0)), $$
wich shows that the limit exists and is zero.
