<s>[INST] Prove that the lim f(x) = 1 using the $\epsilon-\delta$ definition 
Show directly (from the $\epsilon-\delta$ definition ) that 
$$ \lim_{x\to-1}\frac{x^{2}-3}{x-1} = 1$$

Attempt
Using 
$$
\left\lvert \frac{x^{2}-3}{x-1} - 1 \right\rvert 
= \left\lvert \frac{x^{2}-x-2}{x-1} \right\rvert 
= \left\lvert \frac{(x-2)(x+1)}{x-1} \right\rvert
$$
and 
$|x-2|\leq |x - 1 - 1 | \leq |x-1|+|-1| = |x-1|+1 \leq \delta + 1 = 2$ (if $\delta \leq$ 1) 
and 
$|x-1| = |x+1-2| \geq |-2| - |x+1| > 2 - \delta \geq 1$ (if |x+1| < $\delta$)
hence 
$$
\left\lvert \frac{x^{2}-3}{x-1} - 1 \right\rvert 
= \left\lvert \frac{x^{2}-x-2}{x-1} \right\rvert 
= \left\lvert \frac{(x-2)(x+1)}{x-1} \right\rvert 
< 2 \delta \leq \epsilon
$$
so $\delta = \frac{\epsilon}{2}$
and picking $\delta$ = min(1,$\frac{\epsilon}{2}$)
hence 
$$ 
\forall \epsilon>0 \exists \delta 
= \min(1,\frac{\epsilon}{2}) > 0 : \forall 
0 < \lvert x--1\rvert < \delta 
= \min(1,\frac{\epsilon}{2}) \implies 
\left\lvert\frac{x^{2}-3}{x-1} - 1 \right\rvert 
= \left\lvert \frac{x^{2}-x-2}{x-1} \right\rvert 
= \left\lvert \frac{(x-2)(x+1)}{x-1} \right\rvert 
< 2 \delta \leq \epsilon$$
hence
$$ \lim_{x\to-1}(\frac{x^{2}-3}{x-1}) = 1$$
 [/INST] I find it easier to make simplifying assumptions first. You have an 
expression $| { (x-2)(x+1) \over x-1 } |$ which you would like to bound near
$x=-1$. Start by bounding $\delta$ by, say, $1$ (that is $|x+1| < 1$), in which case you have
$|x-1| > 1$, and $|x-2| < 4$. Then you have
$| { (x-2)(x+1) \over x-1 } | \le 4 |x+1|$.
Now choose $\epsilon>0$ and let $\delta = \min (1, {1 \over 4} \epsilon)$ to
get the desired result.
</s>