Suppose $f : I \to \mathbb{R}$ where $I$ is an open interval (say $(a,b)$). Also assume that $f(x) > 0$ for all $x \in I$. Then
If $\lim_{x \to a^+} {1 \over f(x)} = 0$, then $\lim_{x \to a^+} f(x) = \infty$
Note that $a^+$ is the limit from the right.
Here is my go. Does it look correct? I am told that I should pick an arbitrary $N$ at the beginning of the proof but did not understand what that means.
Proof
Suppose $f : (a,b) \to \mathbb{R}$ and $\lim_{x \to a^+} {1 \over f(x)} = 0.$ By definition, this means $\forall \epsilon>0$, $\exists m\in(a,b)$ such that if $a < x < m$, then $|1/f(x) - 0|< \epsilon$.
With $\epsilon>0$, we have $1/f(x) < \epsilon$ whenever $a < x < m$. This implies that if $a < x < m$ then $ f(x) > 1/ \epsilon >0$. It follows that we can find some $N = 1/\epsilon> 0$ such that $f(x) >N$ whenever $a < x < m$. This completes the proof that $\lim_{x \to a^+} f(x) = \infty $}