If $f \to \infty$ as $x\to 0$ then $1/f \to 0$ as $x \to 0$. I want to show that if $f:(0,\infty)$ is a function such that $f(x)\to +\infty$ as $x\to 0$, then $1/f(x) \to 0$ as $x\to 0$.
I'll need to use the epsilon-delta definition of a limit to do this.
I'm wondering whether the following statement is correct:
The above for $f(x)$ is equivalent to saying that $\forall M >0\space \exists N>0$ such that $x > N \implies f(x) > M \space\forall\space x \in (0,\infty)$
If it is correct, how do I get from here to my conclusion about $1/f(x)$?
 A: First, note that since the domain of $f$ is $(0, \infty)$, the limit must be as $x \to 0^+$.  Your given information is that $f(x) \to +\infty$ as $x \to 0^+$, which means:
\begin{equation}\tag{1}
\forall M \exists \delta > 0 \forall x(0 < x < \delta \Rightarrow f(x) > M).
\end{equation}
What you have to prove is that $1/f(x) \to 0$ as $x \to 0^+$, which means:
\begin{equation}\tag{2}
\forall \epsilon > 0 \exists \delta > 0 \forall x(0 < x < \delta \Rightarrow |1/f(x)| < \epsilon).
\end{equation}
Hints for the proof:  First look at your goal (2).  The form of that statement suggests that you should start with "Let $\epsilon$ be an arbitrary positive number," and then you should try to come up with a good choice for $\delta$.  The form of (1) suggests that to use it, you should come up with something to plug in for $M$.  (1) says that no matter what number you plug in for $M$, there will be a corresponding $\delta$ with a certain property.
A: $$\forall M:\exists\,\delta > 0:\forall\,|x|< \delta :f(x) > M$$
is equivalent to
$$\forall M:\exists\,\delta > 0:\forall\,|x|< \delta :\frac1{f(x)}<\frac1M$$
and to
$$\forall\epsilon:\exists\,\delta > 0:\forall\,|x|< \delta :\frac1{f(x)}<\epsilon.$$
A: Let $\epsilon\gt 0$ be given.
Since $\lim_{x\to 0}f(x)=\infty$, for every $N\gt 0$ there exists $\delta\gt 0$ such that $f(x)\gt N$ if $x\lt\delta$.
By the Archimedean ordering of the real numbers, given $\epsilon\gt 0$, we can find $N$ such that $N\epsilon\gt 1\implies \frac{1}{N}\lt\epsilon$.
Then $x\lt\delta\implies f(x)> N\implies\frac{1}{f(x)}\lt\frac{1}{N}\lt\epsilon$
Therefore, $\lim_{x\to 0}\frac{1}{f(x)}=0$
