I don't understand this specific part of proving limits of quadratics

Question

I have no trouble understanding the epsilon-delta definition of a limit and proving linear equations with it. However, I'm having trouble with quadratics, specifically when you reach the part where $\delta=1$. I was following along this youtube video. The guy in the video simply went from $$ |x+5||x-2|<\epsilon $$ to $$ 8|x-2| < \epsilon $$ How does this keep inequality true? His reasoning was that since $|x+5|<8$ when $\delta=1$, we could substitute $|x+5|$ by $8$. But this reasoning makes absolutely zero sense to me, can somebody explain what he was doing here?

Answer 1

To be a touch more precise, we can obtain $|x + 5| < \delta + 7$ for any $\delta,$ so for all $\delta \leq 1$ we have $|x + 5| < 8.$ And because we have $|x - 2| > 0,$ multiplying on both sides we get $|x + 5||x - 2| < 8|x - 2|$ whenever $\delta \leq 1,$ so if we want $|x + 5||x - 2| < \epsilon,$ it is sufficient to make $8|x - 2| < \epsilon.$

It may help to write it out the other way after obtaining our expression for $\delta$: bringing it together, if we have $\delta = \min(1, \frac\epsilon 8),$ then because $|x - 2| < \delta \leq \frac \epsilon 8$ we have $8|x - 2| < \epsilon,$ and because $\delta \leq 1$ we have $|x + 5| < 8,$ so we obtain $|x + 5||x - 2| < 8|x - 2| < \epsilon,$ so for any $\epsilon,$ $|x - 2| < \delta(\epsilon)$ implies that $|f(x) - L| < \epsilon$ as desired.

Answer 2

When he "went" from $|x+5||x-2|<\epsilon$ to $8|x-2| < \epsilon$ by "substitution", he did not mean the two inequalities are equivalent, not even that the 1st one implies the 2nd one.

What he had in mind was: if I already made sure that $|x+5|<8$ (by firstly forcing $x$ to stay between $2-1$ and $2+1$) and if I want $|x+5||x-2|<\epsilon$, I just have to make sure (by some additional restriction on $x$) that $8|x-2| < \epsilon$ (because then, I shall get $|x+5||x-2|<8|x-2|<\epsilon$).