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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?

user14122102
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    Lots of missing equations, words, and — to answer your question — absolute values. – Ted Shifrin Jan 29 '21 at 05:28
  • @user14122102 Your explanation is very difficult to follow. Also, when you multiply a non-negative real number to both sides of an inequality, the sign typically needs to be flipped. – Albert Jan 29 '21 at 05:29
  • There are many solved examples of $\epsilon, \delta $ here (for example this answer) which you should study in order to understand the concept. Also notice that once you start proof based mathematics, the symbols of math take backstage and the usual language (like English) becomes more important. You also need to be able to express your argument or steps in clear language. Without that no one will read your proof (or at least feel rather disinterested). – Paramanand Singh Jan 29 '21 at 06:35

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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 $|x-1||x-7| \lt 7 \cdot \frac \epsilon 7=\epsilon$

Ross Millikan
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  • Sorry but how did you get the $|x−7|<7$? where did the bigger than 7 come from? – user14122102 Jan 29 '21 at 05:49
  • Because we started by saying $\delta \lt 1$, so the least $x$ can be is $0$, so the most $|x-7|$ can be is $7$. We just need to bound it. It would have been true, if less clear where it came from, to say $|x-7| \lt 42$, which would make our result $\delta \lt \frac \epsilon {42}$. That is a fine value as well. We just have to show that some $\delta$ works. We don't have to make it as large as it can be. – Ross Millikan Jan 29 '21 at 14:08