Function Larger than Divergent Function is Divergent

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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 $N_1$.

$\blacksquare$


Linguistic Note

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


Also see