What does a norm $\|x\|$ goes to infinity mean? I am looking into Coercive functions. 
The definition says :
A continuous function $f : \mathbb{R}^n → \mathbb{R}$ is called coercive if
$$\lim_{\|x\| \to \infty} f(x) = + \infty$$


*

*What does a norm to infinity mean?

*I have two functions one of which is coercive and another not. I have trouble deciphering the difference.
$$i)\space g(x) = x_1^6 + 5x_2^4 + x_3 - 3x_1x_2x_3^3; \space x \epsilon \mathbb{R}^3\implies \text{not  coercive}\\
ii)\space h(x) = x_1^4 + x_2^4 - 3x_1^3 + x_2; \space x \epsilon \mathbb{R}^2 \implies \text{coercive}$$

 A: For every $a$ there exists $b$ such that for every $x$ with $\|x\| >b$, we have $f(x)>a$.
A: A norm going to infinity can be seen as the length of a vector going to infinity, i.e. an infinitely long "arrow" starting at the origin.
According to your definition a function is then coercive if the function value also go to infinity, independent of the direction of the vector.
So your second example is coercive in $\mathbb{R}^2$ because for the norm going to infinity, either $x_1$, $x_2$ (or both) should go to infinity. Since both $x_1$ and $x_2$ have positive dominating fourth order terms also $h$ goes to infinity.
Your first example is not coercive in $\mathbb{R}^3$ because if you take $\mathbf{x}=(1,1,R)$ and you let $R$ going to infinity, the norm of $\mathbf{x}$ will go to infinity, but because of the negative dominating third order term, $g$ will not go to infinity.
A: Implicit limits are most easily understood in terms of sequences. Thus, 
\begin{equation}
\lim_{||x||\rightarrow \infty}f(x)=\infty
\end{equation}
is equivalent to saying that for every sequence $(x_n)$ we have
\begin{equation}
||x_n|| \stackrel{n\rightarrow\infty}{\longrightarrow} \infty \quad\Rightarrow \quad f(x_n) \stackrel{n\rightarrow\infty}{\longrightarrow} \infty
\end{equation}
A: A hint for 2.:
"Coercive" means that $f(x)$ is large positive when $\|x\|$ is large. Now one of the two given functions can even be large negative for suitable $x$ far away from the origin. For the other function you have to prove that the highest degree terms in any case outweigh the other terms when $\|x\|$ is large, so that only the contribution of the highest degree terms counts in the limit.
