Calculus: Proving $\lim_a \frac{1}{f}=0$ from $\lim_{a^+} f= -\infty$ and $\lim_{a^-} f= \infty$ 
Suppose that $\displaystyle\lim_{x\to a^-} f(x)=\infty$ and $\displaystyle\lim_{x\to a^+} f(x)=-\infty$. Using only the definitions of limit and infinite limit, prove that
  $$
\lim_{x\to a} \frac{1}{f(x)}=0.
$$

I am having difficulties proving this statement using $\varepsilon$-$\delta$ definitions. Any help and guiding steps would be appreciated. Thank you!
 A: Fix $\varepsilon > 0$.
By definition and the assumptions, you know that there exist (by choosing $A\stackrel{\rm def}{=} \frac{1}{\varepsilon}>0$) $\delta_+,\delta_->0$ such that respectively
$$
f(x) > A \qquad \forall x \in (a-\delta_-,a)\tag{1}
$$
$$
f(x) < -A \qquad \forall x \in (a,a+\delta_+)  \tag{2}
$$
(can you see why? This is true for all $A$, so in particular for any convenient one for us, the one above.)
This in particular implies that
$$
0 < \frac{1}{f(x)} < \frac{1}{A} \qquad \forall x \in (a-\delta_-,a) \tag{3}
$$
$$
0 > \frac{1}{f(x)} > -\frac{1}{A} \qquad \forall x \in (a,a+\delta_+)  \tag{4}
$$
i.e. putting (3) and (4) together,
$$
\left\lvert \frac{1}{f(x)}\right\rvert < \frac{1}{A} \qquad \forall x \in (a-\delta_-,a+\delta_+) \tag{5}.
$$
Now, remember our choice of $A$, and consider $\delta\stackrel{\rm def}{=}\min(\delta_-,\delta_+)$.
A: Let $\varepsilon>0$; by assumption, there are $\delta_1>0$ and $\delta_2>0$ such that,


*

*for $a-\delta_1<x<a$, $f(x)>1/\varepsilon$,

*for $a<x<a+\delta_2$, $f(x)<-1/\varepsilon$.
Set $\delta=\min(\delta_1,\delta_2)$ and suppose $0<|x-a|<\delta$. Try proving that
$$
\left|\frac{1}{f(x)}-0\right|<\varepsilon
$$
which will end the proof.
