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!
 A: 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.
