- #1

thenthanable

- 5

- 0

_{x→0}1/(x

^{2}-1)=-1

0<|x|<δ and |((1/(x

^{2}-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

__(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?)__

**-2<x-1<0 → 0<1-x<2 → |1-x|=|x-1|<2**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?