Undertanding why you restrict delta during limit proofs If you want to prove $\lim_{x\to2}(x^2-3x+1) = -1$
Then let $\epsilon \gt 0$ be given and we want to find a $\delta \gt 0$ such that whenever
$0 \lt|x-2| \lt \delta$ we have $|(x^2-3x+1)+1| \lt \epsilon$
The general method i follow before i actually start the proof is to work backwards from that inequality to find an equivalent inequality showing the range of $x$ around $2$ which satisfies it in terms of $\epsilon$; i.e something like $|x-2| \lt \epsilon/3$ for example.
But with this example you end up with
$|(x-1)||(x-2)|<\epsilon$ 
and have to restrict $\delta$ as the whole point is to get $\delta$ in terms of $\epsilon$. If you restrict $\delta$ to be say $\delta \lt 1$ then i get how this restricts range of $x: 1 \lt x \lt 3$ and i have looked at the solution where they find the upper bound of $|x-1|$ to be $2$. Sub that in and then show the implication but i get lost during this whole part because i can't follow what is going on. 
Could someone help me out, thanks. 
 A: Here is a more general consideration which gives hopefully some insight. You might play it through with your specific example to get a grasp of it:
You want to prove $\lim_{x\to x_0}f(x) = L$ using $\epsilon-\delta$.
So, you start with $|f(x) - L|$ and try to bound it as follows:


*

*$|f(x)-L| \leq \color{blue}{|p(x)|}|x-x_0|$, where $\color{blue}{p(x)}$ is a function defined in a neighbourhood of $x_0$ (maybe after removing $x_0$).


You want to have


*

*$\color{blue}{|p(x)|}|x-x_0| \stackrel{!}{<} \epsilon$
So, you need to control $\color{blue}{|p(x)|}$. Here is the point where the restrictions for $\delta$ come into play. It is often possible to find a bound for $\color{blue}{|p(x)|}$ in a $\color{blue}{\mbox{neighbourhood}}$ of $x_0$:


*

*$\color{blue}{|p(x)| \leq M}$ for $0 < |x-x_0| < \color{blue}{\delta_0}$
Now you get 
$$|f(x)-L| \leq \color{blue}{|p(x)|}|x-x_0| \leq \color{blue}{M}|x-x_0| < \epsilon \mbox{ for } \delta < \min\left(\frac{\epsilon}{\color{blue}{M}} ,\color{blue}{\delta_0} \right)$$
A: You have to use the definition. Let $\epsilon >0$, you want to find $\delta = \delta(\epsilon)>0$ (which depends of $\epsilon$) such that $$ 0<|x-2|<\delta \Rightarrow |x^2 -3x + 1 + 1| < \epsilon.$$
First you do $$ |x^2 - 3x + 1+1|=|x^2-3x + 2| = |(x-2)(x-1)|=|x-2|\cdot|x-1| $$
Then let $\delta = 1$ $$ |x-2|<\delta \Rightarrow |x-2|<1 \Rightarrow -1<x-2<1 \Rightarrow1<x<3 \Rightarrow 0<x-1<2 \Rightarrow|x-1|<2 $$
So that $$ |x-2|\cdot|x-1| < 2 \cdot |x-2| < 2 \delta = \epsilon \iff \delta = \frac{\epsilon}{2} $$
Then, you have to choose $\delta$ = min{1, $\frac{\epsilon}{2} \} $
