Epsilon-delta proof for a limit of a troublesome function

[thenthanable]

lim_x→01/(x²-1)=-1

0<|x|<δ and |((1/(x²-1)) +1)<ε

Working on the second equation, I eventually got
|x|< (ε|x-1||x+1|)^0.5
Suppose I restrict |x|<1
-1<x<1,
then
0<x+1<2 → |x+1|<2
-2<x-1<0 → 0<1-x<2 → |1-x|=|x-1|<2 (Is this step correct? In one of the textbooks, it says that we must always have x>0 which is important to conclude that for example, |x-1|<2 in the context of this question. Unfortunately, I don't see why we must conclude that x>0 because doesn't the modulus get rid of all negative values?)

So my choice for δ works out to be δ=min{1,2(ε)^0.5}

However when I work backwards the problem arises.
Suppose I choose δ=1 → 2(ε)^0.5 > 1 → ε>0.25

When I draw the graph, my δ-range of x=0 starts from the asymptote at x=-1 to the other asymptote x=1, and if this is the case, doesn't that mean the range ε about f(x)=-1 is infinite?

 

[LCKurtz]

You have$$
\left |\frac 1 {x^2-1}+1\right |=\left | \frac {1+x^2-1}{x^2-1}\right |= \left |\frac{x^2}{x^2-1}\right |$$which you want to make small. The numerator will be small as $x\to 0$ but if $x$ gets anywhere near $\pm 1$, that denominator will make the fraction large. So you must keep $x$ away from $\pm 1$. That is easy since you have $x\to 0$. So start by restricting $|x|<\frac 1 2$. Figure out the smallest $|x^2-1|$ can be and overestimate $\frac{x^2}{|x^2-1|}$ by putting that smallest value in the denominator. Then you will be ready to figure out $\delta$ to make it all less than $\epsilon$.