lim 1/(x-3) (epsilon delta)

Question

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

Answer 1

You need to bound $\vert x - 3 \vert$ from below. Note that if $\vert x - 1 \vert < 1$, then $0 < x < 2$. Hence $\vert x - 3\vert \geq 1$. So one constraint you want to put on $\delta$ is $\delta < 1$.

Answer 2

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 $\epsilon > 0$, $$ \left|\frac{1}{x-3}+\frac{1}{2}\right| < \epsilon \mbox{ whenever } |x-1| < \delta=\min(\epsilon,1). $$