Limit using epsilon-delta definition The question is:
Use the $\epsilon - \delta$ relationship  to establish the limit $\lim\limits_{x \to  2} \frac{1}{1-x} =-1$ 
This is my answer: 
Let $f(x)=1/(1-x)$ s.t $|f(x)-L|=|x-2|/|x-1|<\epsilon \iff |x-2|<|x-1|\epsilon$
Let $|x-2|<1 \Rightarrow |x-1|<2$, so the $\max(x-1)=2$.
So $\delta(\epsilon)=\inf(1,2\epsilon)$.
Then just follow the proof for $\delta(\epsilon)$ depending on $\epsilon$ greater or less than $1/2$.
Are my steps ok? Specifically, is choosing $1$ ok? How is this arbitrary $|x-2|<1$ chosen?
Thanks
 A: From by point of view, I think you can proceed instead in the following simple way:
we are seeking for  $\delta$, such that for any $x$ staisfying  $|x-2|<\delta$ we have   $\frac{|x-2|}{|x-1|}<\epsilon$.
Indeed,  for  $|x-2|< \delta$ we may write  $$ -\delta <x-2<\delta $$
 and so  $$ 1-\delta <x-1<1+\delta $$ So if we take $\delta \leq 1$, then  $1-\delta \geq  0$, and so $$ 0\leq 1-\delta <x-1<1+\delta $$ Then
$$ \frac{1}{ 1+\delta} <\frac{1}{x-1} <\frac{ 1}{1-\delta} $$ Thus 
$$  \frac{|x-2|}{|x-1|} < \frac{\delta }{1-\delta}$$
So if we take  $\delta $ so that  $\frac{\delta }{1-\delta}\leq \epsilon$  with   $\delta \leq 1$ then we are done.  More precisely, $\frac{\delta }{1-\delta}\leq \epsilon$ means  $ \delta \leq  \frac{\epsilon}{1+\epsilon}$. So let $\delta= min{1,\frac{\epsilon}{1+\epsilon} }$.
To answer your questions, yes you can add a constraint on $\delta$ as  taking  $\delta \leq 1$, however this constrain must be helpfull . In fact, what  you are missing in your proof is to bound  $|1-x|$ from below, since  we need  $\frac{1}{|x-1|}$ to be bounded from above. So even when you take $\delta \leq 1$ you haven't use it in the right way to prove the boundness of  $\frac{1}{|x-1|} $. Just this was missing.  
A: Well it's formally wrong (or too unclear to express it nicer). And somewhat unclear. Your question is also somewhat unclear (there are many $1$s).
The unclear part is that you introduce $L$ without mentioning that $L=-1$ which makes it not so obvious what happens when you state that $|f(x)-L| = |x-2|/|x-1|$.
The formally wrong is that you should prove that $x<\delta$ implies that $f(x)-L<\epsilon$, but from your reasoning it appear claim implication the other way. This leads to a wrongful estimate of $\delta$ (which you can see if you calculate $f(2-2\epsilon)$ for some small $\epsilon$.
Minor points are that $\max$ means maximum (that has to be taken), it's wrong to say that $\max(x-1)=2$ since $x-1<2$. And $\inf$ means infimum (that need not be taken), normally one doesn't use that with only two elements instead it should be $\min(1,2\epsilon)$.
By calculating $f(2-\delta_{max})$ (above) you can see that the result is slightly more than $-1+\delta_{max}$ which indicates that a estimate of even $\delta = \epsilon$ isn't enough. What you could try is to assume that $\delta = \max(1/4,\epsilon/2)$ instead and work your way backwards. What we need is an estimate of $1/|1-x|$ to do this we can use that $1/1-x = (1+x)/(1-x^2)$ and if $|x-2|<1/4$ you have $x>7/4$ and therefore $1-x^2 < 1-49/16 = -33/16$, so $|1-x^2| > 2$. This can be used to estimate $|f(x)-(-1)|$ if $|x-2|<\delta$:
$$|f(x)-(-1)| = \left|{x-2\over x-1}\right| = \left|{(x-2)(x+1)\over 1-x^2}\right| < |(x-2)(x+1)|/2 = |x-2||x+1|/2$$
Now we have $|x-2|<\delta \le \epsilon/2$ and we have that $7/4 < x+1 < 9/4$ so this means
$$|f(x)-(-1)| < {\epsilon\over2} {9\over 4}{1\over2} = {9\over 16}\epsilon < \epsilon$$
The actual way you find the estimate for $\delta$ is to just limit it to stear clear of the singularity for $x=1$, so you assume that $\delta<1/4$ for example, then you calculate an estimate of $|f(x)-(-1)|$ in an easy form based on $\delta$ then you can see what $\delta$ must be in relation to $\epsilon$, then you see that $\delta<\epsilon/2$ would do and you insert that in your estimate too (making $\delta = \min(1/4, \epsilon/2)$) and update your calculation.
