How to show $f(x_1,x_2) = (x_1-x_2)^4 - 5 x_1 x_2$ is not coercive?

Question

How can you show that $f(x_1,x_2) = (x_1 - x_2)^4 - 5 x_1 x_2$ is not coercive?
I somehow have to show that $\lim_{||x||\to\infty} \neq \infty$.
I tried expressing the function as a function of $x_1^2+x_2^2$, but I couldn't find an expression for which I could deduce the limit. Can someone tell me how to do this?

Answer 1

Since $\underset{x_1 \to \infty}{\lim} f(x_1,x_1) = \underset{x_1 \to \infty}{\lim} -5x^2_1 = -\infty$, we have $\underset{\lVert x \rVert \to \infty}{\lim} f(x_1,x_2) \neq \infty$.

Answer 2

Consider the function values on a circle with radius $r>0$ centered at the origin. At $\sqrt{\frac12}(r,r)$, the function is negative, while at $(r,0)$ the function is positive. So somewhere between those two points the function is $0$. (One could, in theory, find this point, but that's unnecessary and possibly difficult.)

Thus arbitrarily far away from the origin, there are points where the function is $0$. So it can't diverge to infinity as $\|x\|\to\infty$.