Find the limit or prove that the limit does not exist $\lim_{x \to c}x^2 + x + 1,$ for any $c \in R$ 
This is what I tried. 
For $\epsilon > 0$ there exists $\delta >0$ such that $0< \mid x - c \mid < \delta (x \in R)$ => $\mid x^2 + x + 1 - c^2 - c - 1 \mid = \mid x^2 + x - c^2 - c \mid < \epsilon$ 
After that I tried to use triangle inequlity $\mid x^2 - c^2 + x - c \mid \leq \mid x^2 - c^2 \mid + \mid x - c \mid$ 
if $\mid x^2 - c^2 \mid + \mid x - c \mid$ is less then $ \mid x^2 - c^2 + x - c \mid$, I was going to make $\mid x^2 - c^2 \mid + \mid x - c \mid$ to $\mid x + c \mid \mid x- c \mid +\mid x - c\mid$ and then $\mid x - c \mid  \mid x+c\mid +1 \mid$ 
and then use $\mid x - c \mid \mid x+c\mid +1 \mid < \epsilon  $
However, I cannot 
I am stuck now. 
I need help. 

Here is a trick:
Start by choosing $\delta \le |c|+1$, then $|x| \le |c|+|x-c| \le 2(|c|+1)$.
Then $|x^2-c^2| = |x+c||x-c| \le 3 (|c|+1) |x-c|$.
Then $|x^2+x+1 - (c^2+c+1)| \le (3|c|+2)|x-c|$.
Now choose $\delta < \min (|c|, {\epsilon \over 3|c|+2})$, then you will
obain the required bound.
