Proving these three crazy limit implications I have this question:
$$\lim_{x\to p^{+}}f(x) = L \neq0, \lim_{x\to p^{+}}g(x)=0$$
and exists a $r>0$ such that $g(x)\neq0$ for all $x \in (p, p+r)$.
In these conditions, show that:
$$\lim_{x\to p^{+}}\frac{f(x)}{g(x)} = +\infty \mbox{ or } \lim_{x\to p^{+}}\frac{f(x)}{g(x)} = -\infty \mbox{ or } \lim_{x\to p^{+}}\frac{f(x)}{g(x)}\mbox{does not exists} $$
My attempt:
To begin with the proof, I did:
$$\lim_{x\to p^{+}}f(x) = L \implies\forall\epsilon>0, \exists\delta(\epsilon,p)| 0< x-p<\delta\implies |f(x)-L|<\epsilon \tag{1}$$
and:
$$\lim_{x\to p^{+}}g(x) = 0 \implies\forall\epsilon>0, \exists\delta_2(\epsilon,p)| 0< x-p<\delta_1\implies |g(x)|<\epsilon \tag{2}$$
With this, for the case that gives me $+\infty$ I did:
$$\lim_{x\to p^{+}}\frac{f(x)}{g(x)} = +\infty \implies\forall M>0, \exists\delta_3(M,p)| 0< x-p<\delta_3\implies \left|\frac{f(x)}{g(x)}\right|>M$$
then, opening the modules of $(1)$ and $(2)$ and taking $\epsilon=M$:
$$|f(x)-L|<\epsilon \implies -\epsilon<f(x)-L <\epsilon \implies -\epsilon + L < f(x) < \epsilon + L \mbox{ or } -M + L < f(x) < M + L $$
$$|g(x)|<\epsilon \implies -\epsilon <g(x)< \epsilon \mbox{ or } -M < g(x) < M$$
Now, to be cleaner, we have:
$$-M + L < f(x) < M + L \implies M-L > -f(x) > -M -L \tag{3}$$
and 
$$-M < g(x) < M \implies -\frac{1}{M}> \frac{1}{g(x)} >\frac{1}{M}\tag{4}$$
I suspect I can, somehow, get:
$$\left|\frac{f(x)}{g(x)}\right|>M \implies -M>\frac{f(x)}{g(x)}>M$$
by multiplying $3$ and $4$ somehow. I suspect, also, that my way of thinking is completely wrong. But maybe you guys can help me with something. Thank you so much!

The three "or" statements in the "to prove" can basically be summarized as $$\lim_{x\to p^{+}}\frac{f(x)}{g(x)}\quad\text{does not exist}$$ 
This suggests a proof by contradiction. Suppose that the limit does exist and equals $N$. Then we can make the fraction arbitrarily close to $N$. $$\left|\frac{f(x)}{g(x)}\right|<\min \{|N|\pm\epsilon\,\}:=N'$$ (where $N$ is assumed to be nonzero) for all $x\in (p,p+\delta)$ for some $\delta>0$.
But we can also make $f(x)$ arbitrarily close to $L\ne 0$, $|f(x)|>|L|-\epsilon$ for all $x$ in a suitable range, and such that $|L|-\epsilon>0$ is strictly positive. Then we have $$|g(x)|>\frac{|L|-\epsilon}{N'}\qquad(1)$$ for all $x\in (p,p+\delta')$, where $\delta'$ is the minimum of all delta's generated previously.
But the quantity on the right side of $(1)$ is a positive number $c$, so we have shown that $|g(x)|>c$ for $x\in(p,p+\delta')$. This contradicts the fact that $$\lim_{x\to p^+}g(x)=0$$ and the proof is complete.
Edit: The case of $N=0$ will have to be handled separately - this should not be too difficult.
