Proof $ \lim_{x \to 3} \frac{x - 3}{x^{2} - 5x + 6} = 1 $ using $ \epsilon-\delta $ definition

Question

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?

Answer 1

$|\frac{x-3}{x^2-5x+6}-1|<\epsilon$ $ \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 $\theta = min(\frac{\epsilon}{2},\frac{1}{2})$

Answer 2

Consider ${|x-3|}$ is at least less than ${\frac{1}{2}}$. Then we have the set ${|x-3|<\frac{1}{2}}$ and will further restrict it for smaller desired epsilons.

But if ${|x-3|<\frac{1}{2}}$, then we know that ${|x-2| > |x-3|}$. Hence we know that ${\frac{1}{|x-2|}<\frac{1}{|x-3|}<2}$.

As you've written the reduced equation is ${\frac{|x-3|}{|x-2|}=\frac{1}{|x-2|}|x-3|}$. Since ${\frac{1}{|x-2|}<2}$ we have ${\frac{1}{|x-2|}|x-3|<2|x-3|}$.

So if we can make ${2|x-3|<\epsilon}$ then we certainly have ${\frac{|x-3|}{|x-2|}<\epsilon}$. Hence, we need ${|x-3|<\frac{1}{2}\epsilon}$ to also be true.

Finally, taking the intersection of this with our initial condition, we have ${|x-3|<\frac{1}{2}}$ and ${|x-3|<\frac{1}{2}\epsilon}$ ${\implies |x-3|<min(\frac{1}{2}, \frac{1}{2}\epsilon)}$.

Let's go back to the original definition of the epsilon-delta limit. For any desired ${\epsilon > 0}$, we need to provide a bound on ${|x-3|}$ that causes the equation to be less than ${\epsilon}$ away from the limit of 1. This function would be as follows: ${\delta(\epsilon) = \begin{cases} \frac{1}{2}\epsilon & 0\lt \epsilon \lt 1\\ \frac{1}{2} & \epsilon \geq 1 \\ \end{cases} }$