Push Theorem

Theorem
Let $f$ be a real function which is continuous on the open interval $\left({a \,.\,.\, +\infty}\right)$, $a \in \R$, such that:


 * $\displaystyle \lim_{x \to +\infty} \ f \left({x}\right) = +\infty$

Let $g$ be a real function defined on some interval $\left({b \,.\,.\, +\infty}\right)$ such that, for sufficiently large $x$:


 * $\forall x: x \in \left({a \,.\,.\, +\infty}\right) \cap \left({b \,.\,.\, +\infty}\right) \implies g \left({x}\right) > f \left({x}\right)$

Then:


 * $\displaystyle \lim_{x \to +\infty} \ g \left({x}\right) = +\infty$

Proof
By the definition of infinite limits at infinity, the given that $f\left({x}\right) \to +\infty$ is:


 * $\forall M_1 \in \R_{>0}: \exists N_1 \in \R_{>0}: x > N_1 \implies f \left({x}\right) > M_1$

Now, the assertion that $g \left({x}\right) \to +\infty$ is:


 * $\forall M_2 \in \R_{>0}: \exists N_2 \in \R_{>0}: x > N_2 \implies g \left({x}\right) > M_2$

For $N_2$, choose $N_1$.

Linguistic Note
In Dutch, this theorem is called Duwstelling, which translates to Push Theorem.

Also see

 * Comparison Test for Divergence