So I have the following problem:
$\lim_{x\to -1}$ $\frac{x+1}{x^2-1}$ $=\frac{-1}{2}$
From using the defintion $|f(x)-L|<\epsilon$
I got
$$\frac{|1+x|}{2|x-1|}$$
I have control over $|x-a|<\delta$
Which in my case is $|x-(-1)|<\delta$
i.e $$|x+1|<\delta$$
Since I don't have any control over $\frac{1}{2|x-1|}$
I assumed that $\delta \le 1$
Following the procedure for finding the interval for $\frac{1}{2|x-1|}$
I got $$-6<2(x-1)<-2$$
from
$$|x+1|<\delta$$
Now using what I know from $$\frac{|1+x|}{2|x-1|} <\epsilon$$
together with
$$|x-(-1)|<\delta$$
Which got me $$|x+1|<-2\epsilon$$
hence
$$\delta=-2\epsilon$$
Which apperently is wrong, it should've beeen $\delta=2\epsilon$ instead.
What am I doing wrong in my calculations? Thank you in advance!
You have gotten
$$-6 < 2(x-1) < -2$$
Since absolute value is a decreasing function when the domain is the set of negative numbers, this implies that
$$2 < 2|x-1| < 6$$
$$1< |x-1| < 3$$
$$\frac13<\frac{1}{|x-1|}<1$$
$$\frac{|1+x|}{2|x-1|}<\frac{\delta}{2}$$