use of definition of limit to prove lim (x->2) x^2-2x+1 
I thought  I knew how to solve this kind of problems, but I really don't know how to start...
 A: You want to show that if, for any $\epsilon>0$, there exists a $\delta>0$ such that whenever $0<|x-c|<\delta$, it follows that $|f(x)-L|<\epsilon$,  then $\lim\limits_{x\rightarrow c} f(x)=L$.
So for example, on (i), you want $|(x^2-2x+1)-1|<\epsilon$, or $|x||x-2|<\epsilon$.  If you choose delta to be no bigger than $1$, then you can say that $|x|$ will be no bigger than $3$.  So you'd want to choose $\delta=\min(1,\epsilon/3)$.  That way, you have, whenever $0<|x-2|<\delta$, it follows that
$$|x||x-2|<3\delta\le3\frac{\epsilon}{3}$$
$$|x^2-2x|<\epsilon$$
$$|(x^2-2x+1)-1|<\epsilon,$$
as desired.
A: You need to show that for each $ \epsilon >0 $, there exists $\delta >0$ such that if $o<|x-2|<\delta$ then $|(x^{2}-2x+1)-1|=|x^{2}-2x|<\epsilon$. 
Let $ \epsilon >0 $ and $|x|<3$. 
Observe that $|(x^{2}-2x+1)-1|=|x^{2}-2x|=|x||x-2|<3|x-2|$. 
Now choose $\delta =\min\{3,\epsilon/3\}$. Then $\delta>0$.
Suppose $o<|x-2|<\delta$. Then $|x-2|<\epsilon/3$ and we have $|(x^{2}-2x+1)-1|<\epsilon $. 
Therefore $$\lim\limits_{x\to 2}x^{2}-2x+1=1.$$
