Please verify my delta-epsilon limit proof of $\lim_{x\to1} x^2 -6x = -5$

Question

I want to prove that the limit exists using the delta-epsilon limit definition. Please somebody verify my solution. The given problem is $$\lim_{x\to1} x^2 -6x = -5.$$ My solution:

Let $ε>0$. Choose $δ>0$ such that $0<δ<1$ and $0<δ<ε/3$.

If $0<|x-1|< δ$, then $|x^2-6x+5| = |(x-5)(x-1)| = |x-5| |x-1|< |x-5|δ < 3δ < ε$

Scratch work: Assume that $δ≤1$, then $|x-1|<δ<1$ implies that $-1<x-1<1$ and $0<x<2$ so that $$5<|x-5|<3.$$

Answer 1

Please post your next math question to the math.stackexchange page.

Since it is short: The principle of your calculation is sound, up to the factorization. However

|x-5| = |x-1-4| ≤ |x-1|+4 < 5

so that initially you need δ<ε/5.

Answer 2

Your scratch work is good up to this point:

$$5<|x-5|<3.$$

This is a claim that $5 < 3$, which is clearly wrong. In fact, from $0 < x < 2$ you can find that $-5 < x-5 < -3$, but in taking the absolute value you change the sign of all the numbers in these inequalities and therefore you reverse the direction of the inequalities: $a < b$ implies $-b < -a$, and $-5 < x-5 < -3$ implies $3 < -(x-5) < 5$. Since this shows that $-(x-5)$ is necessarily positive, you can then substitute $\lvert x - 5 \rvert$ for $-(x-5)$ in the inequalities:

$$3<|x-5|<5.$$

You therefore have a bound of $5\delta$ where you wrote $3\delta$, and therefore you want $\delta < \varepsilon/5$ rather than $\delta < \varepsilon/3$. (The other conditions, $0 < \delta$ and $\delta < 1$, are OK and are necessary for your argument.)