The theorem is a bit long so I didn't want to include it in the title. Basically, here's what I'm trying to prove:
Let $f$ and $g$ be two functions. Suppose that $f(x)$ is bounded in some deleted neighbourhood of $x_0$ and $g(x) \to \infty$ as $x \to x_0$. Then:
$\lim_{x \to x_0} \frac{f(x)}{g(x)} = 0$
$\lim_{x \to x_0} \frac{g(x)}{f(x)} = \infty$
Proof Attempt:
Let $M>0$. Then, there exists a $\delta_1$ such that:
$0 < |x-x_0| < \delta_1 \implies |f(x)| \leq M$
Since $g(x) \to \infty$ as $x \to x_0$ if and only if $\frac{1}{g(x)} \to 0$ as $x \to x_0$, for any $epsilon > 0$, there is a $\delta_2$ such that:
$0 < |x-x_0| < \delta_2 \implies |\frac{1}{g(x)}| < \frac{\epsilon}{M}$
Define $\delta = \min\{\delta_1,\delta_2\}$. Then:
$0 < |x-x_0| < \delta \implies |\frac{f(x)}{g(x)}| < M \cdot \frac{\epsilon}{M} = \epsilon$
Hence, that proves that $\frac{f(x)}{g(x)} \to 0$ as $x \to x_0$.
Let $h(x) = \frac{f(x)}{g(x)}$. Then:
$\lim_{x \to x_0} h(x) = 0 \iff \lim_{x \to x_0} \frac{1}{h(x)} = \infty$
Hence, $\lim_{x \to x_0} \frac{g(x)}{f(x)} = \infty$.
Alternatively, for any $\epsilon > 0$, there exists a $\delta_1$ such that:
$0 < |x-x_0| < \delta_1 \implies |g(x)| > \epsilon$
For any $M>0$, there exists a $\delta_2 > 0$ such that:
$0 < |x-x_0| < \delta_2 \implies |f(x)| \leq \frac{\epsilon}{M}$
Define $\delta = \min\{\delta_1,\delta_2\}$. Then:
$0 < |x-x_0| < \delta \implies |\frac{g(x)}{f(x)}| > \epsilon \cdot \frac{M}{\epsilon} = M$
That proves that $\lim_{x \to x_0} \frac{g(x)}{f(x)} = \infty$.
Does my argument above work? If it doesn't, why? How can I fix it?