Find the limit or prove that the limit does not exist.
$\lim_{x \to c}x^2+x+1$, for any $c \in R$.
Let $\delta \le |c|+1$. Then, if $|x-c|<\delta,|x-c|<|c|+1 $. Then, we have
$|x|=|x+c-c|\le |x-c|+|c|<2|c|+1$.
In addition , we have
$|x^2-c^2|=|x+c||x-c|\le (|x|+|c)|x-c|<(3|c|+1)|x-c|$.
Then, for given $\varepsilon > 0$, we can find a $\delta< \min\{|c|+2,\frac {\varepsilon}{3|c|+1}\}$.
Then, if $|x-c|<\delta, $ we have
$|x^2+x+1-c^2-c-1|=|x^2-c^2+x-c|\le |x+c||x-c|+|x-c|=|x-c|(|x+c|+1)\le|x-c|(|x|+|c|+1)<|x-c|(3|c|+1)<\delta(3|c|+1)<\varepsilon. $
Therefore, the limit does exist, and the limit is $c^2+c+1.$
I borrow an idea from here. I don't understand some explanations here, so I write in similar but different way. Could you check if the proof is okay?
Thank you in advance.
Yes, this argument looks good to me!
On proof structure, it's typical to start an analysis proof with the line "Let $\epsilon > 0$." and proceed from there, to emphasise that the $\epsilon$ used is arbitrary and cannot depend on anything you choose. On that note, it's often clearer to say Pick $\delta$ instead of Let $\delta$ for similar reasons, you are choosing the $\delta$. Finally, since you have choice of $\delta$, I'd pick it to be a concrete value, $\delta = |c| + 1$. This has the main benefit of ensuring $\delta > 0$ (as sometimes required by the limit definition) and just shows concreteness.
However these are all proof-writing tips, your argument has no issues.