prove that $\displaystyle lim_{x\rightarrow2}\frac{x}{x+1}\neq1$ by epsilon delta

defintion.

we need to prove that their is $\displaystyle \varepsilon>0$ so for every $\displaystyle \delta>0$

their is x $\displaystyle |x-2|<\delta$ and $\displaystyle |\frac{x}{x+1}-1|\geq\varepsilon$

so we are free to choose our epsilon and our x

i choose $\displaystyle \varepsilon=2$

x cannot be equal to -1 because of the denominator.

and we need to choose x for which $\displaystyle |\frac{x}{x+1}-1|\geq2$

$\displaystyle |\frac{x-x-1}{x+1}|\geq2$ -> $\displaystyle |\frac{1}{x+1}|\geq2$ -> $\displaystyle \frac{1}{2}\geq|x+1|$

-> $\displaystyle -\frac{1}{2}\leq x+1\leq\frac{1}{2}$ -> $\displaystyle -\frac{3}{2}\leq x\leq-\frac{1}{2}$

but x=-1 there so we have a problem.

where is the problem in my way???

the prof solved it by $\displaystyle \varepsilon=1/6$

$\displaystyle 2>x>max\{2-\delta/2,1\}$

cant understand why he choose such x