Basically, I have two functions: $$ f(x, y) = \sqrt{(x-y^2)^2 + x^4} $$ and $$ g(x, y) = | (x-y^2)^3 | $$ I need to compare them in a punctured neighborhood of zero. I am a bit stuck here.
Edit: compare in sense that there exists a punctured neighborhood of zero where for all $x$ and $y$ from that neighborhood one function is greater than the other.
I.e. I want to prove that $$ \exists O(0,0):\quad \forall x, y \in O(0,0) \quad |f(x, y)| > |g(x,y)| $$
Let $ w = x - y^2 $.
If $ w > 0 $ and $w < 1$:
$$ f = \sqrt{w^2 + x^4} \ge \sqrt{w^2} = w $$
$$ g = w^3 < w $$
Therefore:
$$ f \ge w > g $$
If w < 0 and | w | < 1
$$ f = \sqrt{w^2 + x^4} \ge \sqrt{w^2} = -w $$
$$ g = | w^3 | < -w $$
Therefore:
$$ f \ge -w > g $$
If $ w = 0 $ and $ x \ne 0 $
$$ f = \sqrt{x^4} = x^2 > 0 $$
$$ g = w^3 = 0 $$
Therefore:
$$ f > 0 = g $$
$ |w| < 1 $ is a pretty big neighborhood, relatively speaking.
Hope this helps.
Ced
If $|(x-y^2)|\le 1$ then $g(x,y)\le \sqrt{(x-y^2)^2}\le f(x,y)$. Can you find $O(0,0)$?
