${f(x) = x^2 + x + 1}$ is continuous on ${\mathbb{R}}$ I need help finding the limit of the function ${f(x) = x^2 + x + 1}$ as ${x\rightarrow c}$ for any ${c \in \mathbb{R}}$. So far I have used the limit definition and triangle inequality up to this point.
$${|(x^2+x+1)-(c^2+c+1)|=|x^2-c^2+x-c|\leq|x^2-c^2|+|x-c|}$$
I do not know how to handle the $|x^2-c^2|$ part within the limit definition.
 A: Let $ c\ge 0 $ and $ x $ such that
$$|x-c|<1$$
Given an $\epsilon>0$, we must find $ \delta>0 $ satisfying
$$|x-c|<\delta \implies |f(x)-f(c)|<\epsilon$$
or
$$|x-c|<\delta \implies |x-c||x+c+1|<\epsilon$$
but
$$|x-c|<1\implies c-1<x<c+1$$
$$\implies 2c<x+c+1<2c+2$$
$$\implies |x+c+1|<2(c+1)$$
thus, to realise the condition $$|f(x)-f(c)|<\epsilon$$
It is sufficient to have
$$|x-c|<1 \text{ and } 2|x-c|(c+1)<\epsilon$$
or
$$|x-c|<1 \text{ and } |x-c|<\frac{\epsilon}{2(c+1)}$$
So, we can take
$$\delta=\min(1,\frac{\epsilon}{2(c+1)})$$
You can do the same if $ c<0.$
A: We fix a point $c$, and show that the given function is continuous in $c$.

*

*Let $\epsilon>0$.


(We search some $\delta=\delta(\epsilon)$ such that the ball centered in $c$ and radius $\delta$ is mapped by $f$ into the ball centered in $f(c)$ and radius $\epsilon$. For this we take a break and swear we will take some $\delta<1$. But then we estimate on some other piece of paper $|x^2-c^2|+|x-c|=|x-c|\cdot|x+c|+|x-c|\le|x-c|\cdot(|x|+|c|+1)\le|x-c|\cdot(|c|+1+|c|+1)$. This estimation is the essence. Now we can...)


*

*Consider / set $\displaystyle \delta$ to be the minimal value between $1$ and  $\frac\epsilon{2(|c|+1)}$.


*Let $x$ be such that $|x-c|< \delta$. Then we have:
$$
\begin{aligned}
|f(x)-f(c)|
&=|(x^2+x+1)-(c^2+c+1)|
\\
&=|(x^2-c^2)+(x-c)|
\\
&\le
|x^2-c^2|+|x-c|
\\
&=|x-c|\cdot|x+c|+|x-c|
\\
&\le|x-c|\cdot(|x|+|c|+1)
\\
&\le|x-c|\cdot(|c|+1+|c|+1)\text{ since $\delta<1$}
\\
&<\delta\cdot(|c|+1+|c|+1)
\\
&\le
\epsilon\ .
\end{aligned}
$$
So we have shown by definition...
$\square$
(Note that the hidden estimation is the receipt to chose $\delta$, and exactly the same estimation finishes the proof.)
