<s>[INST] How do i prove this using precise definition of a limit? $(x^2 - 6x +11) = 6$
lim approaching 1
Given ϵ  > 0
if |x−1|<δ then $|x^2 − 6x + 11 -6|$=$|x^2 - 6x +5|$<ϵ
this gets factored to:
$|x^2 - 6x +5|$ = (x-1)(x-5)<ϵ
This is where i am stuck, can someone explain to me what to do next?
So basically we start with $-1<x-1<1$ = $0<x<2$ for (x-1) and we find the equivalence for (x-5) $-5<x-5<-3$
is that correct?
Just to make sure but is it $-5<x-5<-3$ or $-5<x-5<3$? because if you  add -5 to both sides it becomes -3.
 [/INST] Here I propose the solution to the general case where $x$ approaches $a\in\mathbb{R}$ for the sake of curiosity.
In general, it can be proven that
\begin{align*}
\lim_{x\to a}(x^{2} - 6x + 11) = a^{2} - 6a + 11
\end{align*}
Indeed, let us consider that $|x - a| < \delta$. Then we have that
\begin{align*}
|(x^{2} - 6x + 11) - (a^{2} - 6a +11)| & = |x^{2} - a^{2} - 6(x - a)|\\\\
& = |(x-a)(x+a) - 6(x-a)|\\\\
& = |x-a||x + a - 6 |\\\\
& = |x-a||(x-a) + 2a - 6|\\\\
& \leq |x - a|(|x-a| + 2|a-3|)\\\\
& < \delta^{2} + 2\delta|a-3| = \varepsilon
\end{align*}
Consequently, for every $\varepsilon > 0$, there corresponds $\delta = -|a-3| + \sqrt{|a-3|^{2} + \varepsilon}$ such that
\begin{align*}
|x - a| < \delta \Rightarrow |f(x) - f(a)| < \varepsilon
\end{align*}
where $f(x) = x^{2} - 6x + 11$, and we are done.
Hopefully this helps!
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