Proving the $\lim_\limits{x\to2}\,\, 2x^2-6x+7=3$. I was just wondering if the following proof makes logical sense and is set out in a manor which is easy to read and understand, to a mathematician.
Prove the $\lim_\limits{x\to2}\,\, 2x^2-6x+7=3$.
$\underline {Proof}:$
Let $\epsilon \gt 0$ be given.
$\underline{Observation}:$
Let $f(x) =2x^2-6x+7$.
$|f(x)-\ell|=|(2x^2-6x+7-(3)|=|2||x^2-3x+2|=2\cdot|(x-2)||(x-1)|$
Let a suitable interval for $x$ be $1\lt x\lt 3$. 
$\implies \,\,2|(x-1)|\lt4,$ and $|(x-2)|\lt 1$.
So, $|f(x)-\ell|=2\cdot|(x-2)||(x-1)|\lt 4|(x-2)|.$
Therefore Choose $\,\,\delta=min\{1,\,\,\frac{\epsilon}{4}\}.$
With this choice of $\delta$,  $|f(x)-\ell|=|2||x^2-3x+2|=2\cdot|(x-2)||(x-1)|\lt 4|(x-2)|\lt4(\delta)\lt 4\cdot \frac{\epsilon}{4}=\epsilon,$
Whenever $|(x-2)| \lt \delta.\qquad\blacksquare$
If this is alright, I do not understand a trivial point, if $\,\,\delta=min\{1,\,\,\frac{\epsilon}{4}\},\,\,$ then how am I able to use $\frac{\epsilon}{4}$ to conclude the proof?
Is is that I do not need the statement $\,\,\delta=min\{1,\,\,\frac{\epsilon}{4}\},\,\,$, and any of the 2 may satisfy the requirement for $\delta$?
 A: The reason you need to take $\delta=\min\{1,\epsilon/4\}$ instead of just $\delta=\epsilon/4$ is that the definition of limit requires that the implication
$$|x-2|\lt\delta\implies|f(x)-f(2)|\lt\epsilon$$
be true for all $\epsilon$.  (That is, the definition says "for all $\epsilon$ there exists a $\delta$ such that....")  The statement
$$|x-2|\lt{\epsilon\over4}\implies|f(x)-f(2)|\lt\epsilon$$
is not true for all $\epsilon$.  For example, let $\epsilon=100$ and $x=10$.  Then $|x-2|=|10-2|=8\lt25=\epsilon/4$ but $|f(x)-f(2)|=|f(10)-f(2)|=|147-3|=144\not\lt\epsilon$.
Now of course the existence and value of a limit ultimately only care about small $\epsilon$'s.  But in writing a proof, it's important to adhere to strict logical rigor.
A: Let $f(x)=2x^2-6x+7$ and $\epsilon>0$.
$|f(x)-f(2)|=2|(x-1)(x-2)|$
as $x$ is near $2$, we can assume that
$|x-2|<\color{red}{1},\;$ for example. which gives
$1<x<3$ or $0<x-1<2$ and $|x-1|<2$.
thus
$|x-2|<inf(\color{red}{1},\frac{\epsilon}{4})$
$\implies$
$|x-1|<2\;\; $and$\;\;|x-2|<\frac{\epsilon}{4}$
$\implies$
$$|f(x)-f(2)|=2|x-1||x-2|$$
$$<2.2.\frac{\epsilon}{4}=\epsilon.$$
therefore, $\lim_{x\to 2}f(x)=3$.
A: For technical reasons you have to choose $\delta=min\{1,\,\,\frac{\epsilon}{4}\}$ to make sure the step "Let a suitable interval for $x$ be $1<x<3$." won't fail. Because otherwise the condition $|x-2| < \delta$ wouldn't ensure that $x\in(1,3)$ if $\delta$ is greater then 1.
Otherwise the $\varepsilon$ is usually very small so that the equation $\delta = min\{1,\,\,\frac{\epsilon}{4}\} = \frac{\epsilon}{4}$ will hold...
