prove that $\lim_{x \to 1}\frac{x^{2}-x+1}{x+1}=\frac{1}{2}$ Please check my proof 
Let $\epsilon >0$ and $\delta >0$
$$0<x<\delta \rightarrow \frac{x^{2}-x+1}{x+1}-\frac{1}{2}<\epsilon $$
$$\frac{2x^{2}-2x+2-x+1}{2x+2}<\epsilon $$
$$\frac{2x^{2}-3x+1}{2x+2}<\epsilon $$
since   $\frac{2x^2-3x+1}{2x+2} <\frac{2x^{2}-3x+2}{2}$
then $\frac{2x^{2}-3x+1}{2}< \epsilon $
$2x^{2}-3x+1<2\epsilon $
choose $2\epsilon =\delta $
then $\frac{2x^{2}-3x+1}{2}<\frac{2\epsilon }{2}=\epsilon $
by transitivity of inequality
$\frac{2x^{2}-3x+1}{2x+2}<\epsilon $   

This function is defined at continuous at x = 1. 
(It would be a different problem is we were looking for the limit as x approaches -1)
Unless your professor is demanding epsilon-delta proofs for the practice, you know that the limit of a defined continuous f as x approaches 1 is 
$f(1) = \frac{1^2 - 1 + 1}{1 + 1} = \frac{1}{2}$
(It is OK to do epsilon-delta proofs for practice if the professor thinks you need more practice; just not necessary here.)
