Is this $\epsilon$-$\delta$ limit correct? I have to show that $\lim_{x\to2}x^2+x+1=7$ with the $\epsilon$-$\delta$ definition of limit. Here is how I have done it: $\left\lvert x^2+x-6\right\rvert=\lvert x−2\rvert\lvert x+3\rvert<\epsilon$ and since we are close to $2$, we can assume that the $\delta$-neighborhood of $c=2$ must be have a radius of max $\delta=1$ which implies that : $\lvert x-2\rvert < \frac{\epsilon}{6}$.
We now choose $\delta = \min\{3,\frac{\epsilon}{6}\}$  and we can conclude that if $\lvert x−2\rvert<\delta$, it follows that $\left\lvert x^2+x+1-7\right\rvert < 6\frac{\epsilon}{6}=\epsilon$. Is this calculation correct? Do I miss something? Or some details?
 A: It really is as simple as you were making it, although there are mistakes and it is badly written. You correctly realized that it is important to bound one of $|x-2|, |x+3|$, namely $|x-2|$.
It is standard to start with $\delta := 1$ as this gives us more information to work with for small $\epsilon$, in particular when we have $0 < \epsilon < 1$.
Remember that we are considering the limit as $x$ approaches $2$, so we need to first look at how to restrict $x$, sometimes you must do this and in this case in particular, because we need to add $5$ for $|x+3|$ while still leaving it bounded in some way (otherwise $x$ could just be negative number where $|x|$ is very large).
That being said, we can restrict $x$ around $2$ such that $1 < x < 3$ so that $|x-2| < 1:\delta_{1} $ and $4 < x+3 <6 \implies |x+3| < 6$.
Hence, if we choose $\delta_{1} := 1$, we see what our other term is bounded by.
Now we can choose $\delta_{2}$ so that $|x-2| < \delta_{2} = \frac{\epsilon}{6}.$
Therefore, for some arbitrary $\epsilon > 0$ if we take $\delta:=\min\{\delta_{1},\delta_{2}\}$, then $|x^{2}+x-6| =|x-2||x+3| < \frac{\epsilon}{6}\cdot 6 = \epsilon$.
A: Normally, before or while I present an answer, I am supposed to respond to the OP's question(s) and point out any errors or omissions.  I simply can't do that here.  I have to agree with Ted Shifrin's comment.
Let $f(x) = x^2 + x + 1.$
To Prove:
$\forall \epsilon > 0 ~\exists ~\delta > 0~$ such that 
$|f(x) - 7| < \epsilon~$ whenever $~0 < |x - 2| < \delta.$
I will present this as a full solution, showing how you craft the relationship between $\delta$ and $\epsilon$.
Suppose that $(2-\delta) < x < (2 + \delta)$.  Then

*

*$4 - 4\delta + \delta^2 < x^2 < 4 + 4\delta + \delta^2.$


*$2 - \delta < x < 2 + \delta.$
Therefore, $7 - 5\delta + \delta^2 < f(x) < 7 + 5\delta + \delta^2.$
The first thing to do is establish that one of the constraints that will be imposed on $\delta$ is that $\delta < 1.$
This will guarantee that $0 < \delta^2 < \delta.$
With this constraint imposed,
$$7 - 5\delta < f(x) < 7 + 6\delta.$$
This means that for any $0 < \delta < 1,$ if $\epsilon$ happens to be greater than $6\delta$, then
$$|f(x) - 7| < \epsilon.$$
So one easy solution is (for example) set $\delta = \min\left[0.9, (\epsilon/10)\right].$
A: We put $f(x) = x^2 +x+1$
We can prove that :
$|f(x) - l|<\delta $ $ \Rightarrow $ $ |x-a|<\alpha $
$\alpha , \delta > 0$
$|f(x) - l|= |x^2 +x+1-7|=|x^2 +x-6|=|x-2||x+3|$
$|f(x) - l|<\delta$
$\Rightarrow $ $|x-2||x+3|<\delta$
Suppose $x\in [\frac{3}{2}, \frac{5}{2}] $
$\Rightarrow $$\frac{9}{2}\leq x+3\leq\frac{11}{2}$
$\Rightarrow $$|x+3|\leq \frac {11}{2} $
$\Rightarrow $$|x-2||x+3|\leq \frac{11}{2} |x-2|$
we know that :
$|x-2||x+3|<\delta$
So:
$\frac{11}{2} |x-2|<\delta$
$\Rightarrow $ $|x-2|<\frac{2} {11}\delta$
We put $\alpha=\frac{2\delta} {11}$
Finally :
After the definition of limite we proved $\lim_{x\to 2} f(x) =7$
