Showing coercivity of a function I am well attuned to the definition for a function to be coerce,
which is that $\lim_{\|x\| \to \infty}f(x) = \infty$
ie the values of $f$ go to infinity as the norm goes to infinity.
So 
Ex.1 $f(x_1,x_2) = x_1^4 + x_2^4 - 3x_1x_2$
I thought this was immediately obvious because 
$x_1^4$ and $x_2^4$ grow faster than the linear terms we have.
 Would this be considered a sufficient proof? 
I am unsure in particular how taking the limit would be shown.
Additionally, I don't see why functions like
Ex.2 $f:\mathbb{R}^n \to \mathbb{R}, f(x) = a^tx$ isn't coercive,
or
Ex.3 $f(x_1,x_2) = x_1^2 + x_2^2 - 2x_1x_2$ isn't coercive.
Since they are both unbounded, doesn't taking the norm of $x \to \infty$ show that these functions also approach infinity?

Ex.1 No, your answer is not rigorous. It is true, but you need to prove it. My suggestion is to show that $$\lim_{\|(x_1,x_2)\| \to +\infty} \frac{x_1 x_2}{x_1^4+x_2^4}=0.$$
Ex.2 If $x \perp a$, then $f(x)=0$. And since on the subspace $\{a\}^\perp$ there are vectors of arbitrarily large norm...
Ex. 3 If $x_1=x_2=t$, then $f(t,t)=0$ for any $t>0$. Letting $t \to +\infty$...
