So i am trying to practice questions about Delta-Epsilon and i got stuck on this one here, $x^2-8x+14=7$ as x approaches 1.
i know that $|x-1|$<$d$ implies that:
$x^2-8x+14=7$
$x^2 -8x+7$
$(x-1)(x-7)$
So if the absolute value of x-1 is less than 1 then $-1<x-1<1$ and so does that mean that $(x-7)$ is equal to $-7<x-7<-5$? Because if that is true then doesnt it mean that $-5|x-1|<ϵ$
$|x-1|<ϵ/-5$
So how can we have a negative in epsilon? can anyone explain where i got it wrong?
Start with what you are trying to satisfy. Given an $\epsilon \gt 0$ you want to ensure that $|x^2-8x+7| \lt \epsilon$. It makes good sense to factor that and get $|(x-7)(x-1)| \lt \epsilon$ but you forgot the absolute value signs. Now note that you can split apart the absolute value to get $|x-7||x-1| \lt \epsilon$ If you start by stating that $\delta$ will be less than $1$ you can guarantee that $|x-7| \lt 7$ so you can say $\delta =\min(1, \frac \epsilon 7)$ is guaranteed to work. All the steps are reversible, but you should go in the other order.
Given $\epsilon,$ I choose $\delta$ to be $\min(1, \frac \epsilon 7)$. If $|x-1| \lt \delta, |x-7| \lt 7$ so $|x-1||x-7| \lt 7 \cdot \frac \epsilon 7=\epsilon$