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 $x$:


 * $g \left({x}\right) > f \left({x}\right)$

Then:


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

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

By the definition of infinite limits at infinity, this means:


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

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


 * $\forall M_1 \in \R_{>0}: \exists N_2 \in \R_{>0}: x > N_2 \implies g \left({x}\right) > M_1$

By the premise that $g \left({x}\right) > f \left({x}\right)$ for sufficiently large $x$, there is an $N$ such that:


 * $x > N \implies g \left({x}\right) > f \left({x}\right)$

Now, given $M_1$, let $N_2$ be greater than both $N_1$ and $N$.

Then since $N_2 > N$ and $N_2 > N_1$ respectively:


 * $g \left({x}\right) > f \left({x}\right) > M_1$

whence:


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

Note on Terminology
The author of this page has not found a name for this theorem in any English source.

Push Theorem is the translation of the Dutch name of this theorem, Duwstelling.

Also see

 * Comparison Test for Divergence