<s>[INST] lim 1/(x-3) (epsilon delta) I want to prove that  $\lim_{x\to 1}$   $\frac{1}{x-3}=-\frac{1}{2}$
So I started:
$|\frac{1}{x-3} + \frac{1}{2}|$=$\frac{|x-1|}{2|x-3|}$<$\epsilon$
My problem is to bound $\frac{1}{|x-3|}$ from $|x-1|$
Thanks, for any help 
 [/INST] Start by restricting $|x-1| \le 1$. Then $-1 \le x-1 \le 1$ gives
$$
                    x-3=(x-1)-2 \\
                    -1-2 \le x-3 \le 1-2 \\
                      -3 \le x-3 \le -1 \\
                         |x-3| \ge 1.
$$
Then, whenever $|x-1| \le 1$,
$$
           \left|\frac{1}{x-3}+\frac{1}{2}\right|=\left|\frac{x-1}{x-3}\right|
   \le |x-1|.
$$
Therefore, for any $\epsilon > 0$,
$$
       \left|\frac{1}{x-3}+\frac{1}{2}\right| < \epsilon
    \mbox{ whenever } |x-1| < \delta=\min(\epsilon,1).
$$
</s>