root of an odd degree polynomial

Im asked to prove that a polynomial of an odd degree has a root. Im going to use IVT to prove this, but I'm wondering if I can use the assumption that an odd degree polynomial has either two cases:

$$ \lim_{x\to\infty}h(x)=+\infty \\ \lim_{x\to-\infty}h(x)=-\infty \\ $$ $$and$$ $$ \lim_{x\to\infty}h(x)=-\infty \\ \lim_{x\to-\infty}h(x)=+\infty \\ $$ Then by this there exists numbers $a,b$ such that $$ h(a)<0<h(b) $$ $$or$$ $$ h(b)<0<h(a) $$ Which I think would prove that odd degreed $h(x)$ has at least a root. Or is there a more convincing way to show that there are numbers in $h(X)$ greater than zero and less than zero?

edit: (sorry about the bad limit notation)

This is a pretty efficient method.

If it's available, one could also appeal to the statement that any polynomial over $\Bbb R$ factors over $\Bbb R$ as a product of linear and quadratic polynomials. In particular, if the degree of a polynomial $p$ is odd, it must have at least one linear factor, and the root of the linear factor is a root of $p$.