Is the following function coercive? Can anyone help Is the following function coercive?
My professor wrote:

In the function $f(x_1, x_2) = x_1^3+x_2^2+3$, we
  notice that the leading term $x_1$. is odd so it can’t be coercive. To
  prove this we consider the restriction to the x-axis ((x, y) = (t, 0))
  so we have $φ(t) = t^3 + 3$, this restriction is such that:
$\lim_{t \to +∞} φ(t) =\lim_{t \to +∞} t^3 + 3 = +∞$ and  $\lim_{t \to
> -∞} φ(t) = \lim_{t \to -∞} t^3 + 3 = −∞$ so the function can’t be coercive.

I do not understand why the fact that the restriction is such that
$\lim_{t \to +∞} φ(t) =\lim_{x \to +∞} t^3 + 3 = +∞$ and 
$\lim_{t \to -∞} φ(t) = \lim_{t \to -∞} t^3 + 3 = −∞$
Implies that the function is not coercive. Is this because this implies that the function is not bounded?
 A: 
Here we have a function $f$
\begin{align*}
&f:\mathbb{R}^2\to\mathbb{R}\\
&f(x_1,x_2)=x_1^3+x_2^2+3\tag{1}
\end{align*}
  from  which  we want to show that it is not coercive. In order to show that $f$ is not coercive a single counter-example is already sufficient. 

Coercive means that  whenever we  consider $x$ with $||x||\to +\infty$ then we have $f(x)\to +\infty$. When analysing $f(x_1,x_2)$ in (1) we see that


*

*when we set $x_2=0$ in $f$ we have $f(x_1,0)=x_1^3+3$

*$\lim_{x_1\to -\infty}f(x_1,0)=\lim_{x_1\to-\infty}x_1^3+3=-\infty$.

Note that this is all we need for our counter example, since $x=(x_1,0)$ with $\lim_{x_1\to -\infty}f(x_1,0)=-\infty$ implies that $\lim_{||(x_1,0)||\to\infty}f(x_1,0)=\color{blue}{-\infty}$ showing that $f$ is not coercive.

In order to make this counter-example even more obvious we can extract this special case as function by its own and consider a restriction $\phi$ with
\begin{align*}
&\phi:\mathbb{R}\to\mathbb{R}\\
&\phi(t)=t^3+3\\
\end{align*}
Note that $\phi(t)$ corresponds to $f(t,0)$. So, we argue as above by taking $t$ with $t\to-\infty$ showing that $\phi(t)$ is not coercive, since it implies that $\lim_{||t||\to\infty}\phi(t)=-\infty$.
