I need to prove that as x-> + infinity, (x+1)/(x-1)= 1
With epsilon delta definitions... this is so confusing, can anyone help?
What you want to show is that for any $\displaystyle \epsilon > 0$ there is an $\displaystyle N > 0$ such that
$\displaystyle \left| \frac{x+1}{x-1} - 1\right| < \epsilon$ when $\displaystyle x > N$
Now start with the left
$\displaystyle \left| \frac{x+1}{x-1} - 1\right| < \epsilon$
$\displaystyle \left| \frac{2}{x-1}\right| < \epsilon$
$\displaystyle \left| \frac{x-1}{2} \right| > \frac{1}{\epsilon}$
$\displaystyle \left| x - 1\right| > \frac{2}{\epsilon}$
or
$\displaystyle x-1 >\frac{2}{\epsilon}$ (x is very large)
or
$\displaystyle x > \frac{2}{\epsilon} + 1$
This is your N, i.e. $\displaystyle N = \frac{2}{\epsilon} + 1$
Then work the steps in reverse.