<s>[INST] Showing $\lim_{x \to2^-}\frac{1}{x-2} = -\infty$ using the definition of the limit 
Show
  $$\lim_{x \to2^-}\frac{1}{x-2} = -\infty $$

My answer: 
For all $B<0,   \exists \delta >0 $ such that $\frac{1}{x-2} < B$ always that $-\delta < x-2 < \delta$, "in this part I don't understand because write $-\delta < x-2 < 0$, because approach for the left not necessary is for negatives values, not it sure??? I really need some explanation in this part"
So, I continue my answer with: 
Looking in inequality between B, we have: 
\begin{align}
\frac{1}{x-2} < B\\
x-2 > \frac{1}{B}\\
x> \frac{1}{B}+2
\end{align}
Then, We choose $\delta=\frac{1}{B}+2$
Like this, $\frac{1}{x-2} < B$ always that $-\delta < x-2 < 0.$
 [/INST] Remember we are assuming that $x < 2$ as $x\to 2^-$.
$-\delta < x - 2 < \delta;$ and $x < 2 \iff $
$-\delta < x-2 < 0 \iff$
$0 < 2-x < \delta \iff$
$\frac 1{x-2} > \frac 1{\delta} \iff$
$\frac 1{2-x} < -\frac 1{\delta}$
So if we set $-\frac 1{\delta} = B$ or in other words if we set $\delta = -\frac 1B$ we have our proof:
If $|x-2| < \delta = -\frac 1B$ then 
$\frac 1B < x-2 < -\frac 1B$ and $x < 2$ so
$\frac 1B < x-2 < 0$ so
$0 < 2-x < -\frac 1B$ so
$ \frac 1{2-x} > -B > 0$ so 
$\frac 1{x-2} < B < 0$.
</s>