Proof limit of function using Cauchy condition: $\lim_{x\to0}\frac{x+1}{x-1}$

Question

I have to calculate limit at 0 of that function:

$\lim_{x\to0}\frac{x+1}{x-1}$

I have problem to find correct $\delta$ I've already done this:

$0\leq|\frac{x+1}{x-1}+1|<\epsilon$

$0\leq|\frac{x+1+x-1}{x-1}|<\epsilon$

$0\leq|\frac{2x}{x-1}|<\epsilon$

$0\leq2|\frac{x}{x-1}|<\epsilon$ I have no ideas what should I do next I know that

$0\leq|x|<\delta$

But what next? How can I show that $\delta$ exists for all $\epsilon$?

Answer 1

Proving for $0<\epsilon<1$ would suffice.

$|x|<\epsilon \implies 0<(1-\epsilon)<(1-x)<(1+\epsilon) \implies 0<(1-\epsilon)<|x-1|<(1+\epsilon) \implies \frac{1}{|x-1|}<\frac{1}{1-\epsilon}$ Here we just exploit the fact that $\frac{1}{|x-1|}$ is bounded for $|x|<\epsilon <1$. Therefore we have $|\frac{x+1}{x-1}+1|=\frac{|2x|}{|x-1|}<\frac{2|x|}{(1-\epsilon)}$. Then we have $\frac{2|x|}{(1-\epsilon)}<\epsilon$ iff $|x|<\frac{\epsilon(1-\epsilon)}{2}$. So we take $\delta=\frac{\epsilon(1-\epsilon)}{2}$. And clearly this $\delta$ is less than $\epsilon$ so that the inequality $\frac{1}{|x-1|}<\frac{1}{1-\epsilon}$ holds.

Answer 2

is that way correct? $f(x_1)=-1+e\rightarrow \frac{x_1+1}{x_1-1}=-1+e \rightarrow (x_1+1)=(-1+e)(x_1-1)\rightarrow(x_1+1)=(-x_1+1+ex_1-e)\rightarrow (2x_1-ex_1)=-e\rightarrow x_1=\frac{-\epsilon}{2-\epsilon}$

same way for $f(x_2)=-1-e$

and than i will chose lesser absolute value and this will be my $\delta$?