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In many calculus exercises we encounter limits like $\lim_{x\rightarrow \infty} 1/f(x)$ where $f(x)$ is some unbounded and strictly increasing function of $x$.

I'm wondering if there's a way to prove that the above limit is $0$ for any such $f(x)$. That is, we want to show that for all $\epsilon>0$ there is an $M= M(\epsilon)$ such that $\forall x>M$ we have $|1/f(x)|<\epsilon$.

The tricky part seems to be how to precisely define $M(\epsilon)$ as its value depends on the actual growth of $f(x)$, which we don't know.

So is it possible to prove the above statement without knowing $f(x)$?

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up vote 1 down vote accepted

Yes. $f$ being unbounded gives you $$ \forall L \ge 0 \; \exists M \in \mathbb R : f(M) \ge L $$ $f$ being increasing gives from this $$ \forall L \ge 0 \; \exists M \in \mathbb R \; \forall x \ge M : f(M) \ge L $$ Now let $\epsilon > 0$ be given, choose $M$ arcording to $L = \frac 1\epsilon$, then for all $x \ge M$ we have $f(x) \ge 1/\epsilon$, that is $0 \le \frac 1{f(x)} \le \epsilon$. This proves $\lim_{x\to\infty} f(x) = 0$.

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Should that say $\lim_{x\to\infty}\frac{1}{f(x)}=0$ at the end? – yunone Nov 13 '12 at 22:28

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