Showing $\lim_{x \to2^-}\frac{1}{x-2} = -\infty$ using the definition of the limit

Question

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 $-\delta < x-2 < 0.$

Answer 1

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

$\frac 1{x-2} < B < 0$.

Answer 2

Let $M<0$. Suppose $x<2$. Set $\delta=-\frac{1}{M}$.

Therefore $$-\delta<x-2<0\implies \frac{1}{x-2}<M.$$