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 $   

There was an error in the second line of the development.
Note that 
$$\begin{align}
\left|\frac{x^2-x+1}{x+1}-\frac12\right|&=\left|\frac{2x^2-3x+1}{2(x+1)}\right|\\\\
&=\left|\frac{(2x-1)(x-1)}{2(x+1)}\right|
\end{align}$$
Now, bound $|x-1|$ by something somewhat arbitrary such as $0<x<2$.  Then $1<x+1<3$ and $-1<2x-1<3$ so that $\frac{|2x-1|}{2(x+1)}<\frac32$ and for $|x-1|<1$
$$\begin{align}
\left|\frac{x^2-x+1}{x+1}-\frac12\right|&=\frac32 |x-1|\\\\
&<\epsilon
\end{align}$$
whenever $|x-1|<\delta=\min\left(1,\frac{2}{3}\epsilon\right)$.  And we are done!
