<s>[INST] Proof $ \lim_{x \to 3} \frac{x - 3}{x^{2} - 5x + 6} = 1 $ using $ \epsilon-\delta $ definition I'm trying to prove that $ \lim_{x \to 3} \frac{x - 3}{x^{2} - 5x + 6} = 1 $ using $ \epsilon-\delta $ definition. Here is my proof so far:
My effort: Let $ \epsilon \gt 0 $, Choose $ \delta = min\{1, \}.$ Suppose $ 0 \lt \left|x - 3\right| < \delta.$
Check: $ \left|\frac{x-3}{x^2 - 5x + 6} - 1\right| = \left|\frac{x - 3}{\left(x - 2\right)\left(x - 3\right)} - 1\right| = \left|\frac{1}{x-2} - 1\right| = \left|\frac{1 - x + 2}{x - 2}\right| = \left|\frac{-x + 3}{x - 2}\right| = \frac{\left|x-3\right|}{\left|x-2\right|}.$
I want to find a lower bound $ M \gt 0$ such that $ \left|x - 2\right| \ge M $ so that I can write $ \frac{\left|x - 3\right|}{\left|x - 2\right|} \le \frac{\left|x -3\right|}{M}$.
I know that $ -1 \lt x-3 \lt 1 $ and therefore $ 0 \lt x - 2 \lt 2 $, but I can't choose $ 0 $ to be the lower bound because it needs to be positive. What can I do?
 [/INST] $|\frac{x-3}{x^2-5x+6}-1|<\epsilon$
$ \Leftrightarrow$ $\frac{|x-3||x-3|}{|x-3||x-2|}<\epsilon$
$ \Leftrightarrow$$\frac{|x-3|}{|x-2|}<\epsilon$
let's put $|x-3|<\frac{1}{2}$ so $\frac{1}{|x-2|}<2$,so finally;
$|x-3|<\frac{\epsilon}{2}$ and $|x-3|<\frac{1}{2}$(our assumption),so we can put $\theta = min(\frac{\epsilon}{2},\frac{1}{2})$
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