Is this proof of the limit $\lim_{x \to 2}(x^2 - 6x) = -8$ legitimate? I am presented with the following task:
Use the formal definition of the limit to prove the limit
$$\lim_{x \to 2}(x^2 - 6x) = -8$$
The way I solved it is very different from the solution presented by the answer-sheet to the exam, thus I am wondering if my proof is legitimate:
Let $|x - 2| < \delta$. From simplicity, bound delta so that $\delta < 1$. Thus we have that $|x-2| < 1$. Now, let $|x^2 - 6x + 8| < \epsilon$. Thus we have that $|x-2||x-4| < \epsilon$. We already have that $|x-2| < 1$, which implies $|x - 2 - 2| < 1 + 2$ (please tell me if my absolute-value logic is off), therefore $|x-4|<3$. This implies $|x-2||x-4|<\epsilon<3\delta$, there an appropiate value for $\delta$ would be $\delta = \frac{\epsilon}{3}$
 A: Your statement involving the absolute value is fine. I don't see why you said $\epsilon$ is less than $3 \delta$. The point behind the proof is, given an arbitrary $\epsilon$, to produce a number $\delta$ satisfying the required condition. In your argument above you started out with $\delta$ (What is that?) and only at the end did you say that $\delta$ should be $\epsilon / 3$. While your work is a good way of finding $\delta$, the heart of the proof is to show that your value of $\delta$ works. In a sense, your work above is scratch work. A complete proof would go something like: Let $\epsilon > 0$ and let $\delta <$ min$\{1, \epsilon/3\}$. Then if $|x-2| < \delta$, we have $|x-2| < \delta < 1$, so $|x-4| = |x-2 + (-2)| \leq |x-2| + |-2| < 3$, so $|x^2-6x+8| < 3\delta < \epsilon$. Therefore $\lim_{x \to 2}(x^2 - 6x) = -8.$
A: Let $\epsilon > 0$
$|x^2 - 6x + 8| = |x-2||x-4|$
Now you want a bound for $|x-4|$, so you take $|x-2|<1$
$\therefore |x-4|= |x-2-4+2| \le |x-2|+|-2| < 1 + 2 < 3$
(using triangle inequality $|x+y| \le |x|+|y|$)
$\implies |x-2||x-4|<3|x-2|< \epsilon$
$\therefore |x-2|< \frac{\epsilon}{3}$
Take $\delta = min \lbrace 1, \frac{\epsilon}{3} \rbrace$ and you are done.
