Understanding the Definition of Coercive Functions

Question

A function $f$ defined on $\mathbb{R}^n$ is said to be coercive if $$\lim_{\|\vec{x}\|\rightarrow \infty}f(\vec{x})=+\infty.$$ I do not understand the idea of taking the limit as the norm approaches infinity. Consider an example using the function $$f(\vec{x})=(x^4+y^4)\left(1-\frac{4xy}{x^4+y^4}\right).$$ As $\|\vec{x}\|\rightarrow\infty, \frac{4xy}{x^4+y^4}\rightarrow0.$ I do not understand this. Do we think of $\|\vec{x}\|\rightarrow\infty$ in the sense that $x\rightarrow\infty$ and $y\rightarrow\infty$? Some intuition would be most helpful.

Answer 1

Proof of the fact that $\frac {4xy} {x^{4}+y^{4}} \to 0$ as ${x^{2}+y^{2}} \to \infty$: Use the fact that $2xy \leq x^{2}+y^{2}$ and consider the cases $|x| \leq |y|$ and $|x| >|y|$. In the first case $\frac {x^{2}+y^{2}} {x^{4}+y^{4}} \leq \frac 2 {y^{2}}$. Second case is similar.

Answer 2

You have to expand what what the notation $\lim\limits_{||\vec x||\to\infty}f(\vec x)=+\infty$ means: you get $$ \forall A\in\mathbb{R}, \exists B>0 \text{ s.t. }\forall \vec x\in\mathbb{R}^n,||\vec x||>B\implies f(\vec x)>A,$$ Also, if you expand the notation $\lim\limits_{||\vec x||\to\infty}f(\vec x)=l\in\mathbb{R}$, you get $$ \forall \varepsilon>0, \exists B>0 \text{ s.t. }\forall \vec x\in\mathbb{R}^n,||\vec x||>B\implies |f(\vec x)-l|<\varepsilon.$$