Proving that $\lim_{x\to 2}(x^2+x-1)=5$ using an $\epsilon$-$\delta$ argument I'm just working on the construction portion at the moment for
$\lim_{x\to 2}(x^2+x-1)=5$.
Then let $\epsilon > 0$ and $|x^2+x-1-5| < \epsilon \iff 
|x^2+x-6|<\epsilon$.
Then I write $|x-2||x+3| < \epsilon$.
I have no idea what to say after this point, my professor said to restrict one of the absolute value's but I'm still not sure what he's getting at.
 A: Informally, if $x$ is near $2$, you know $x+3 \gt 0$ so you can just remove the absolute value bars and $x+3$ won't be too far from $5$.  To be careful about it, you can just say that you will require that $\delta \lt 1$ (or anything reasonably small) which is enough to guarantee that $4 \lt x+3 \lt 6$.  Then do your computations that will result in a $\delta$ that depends on the $\epsilon$ you are given.  Normally $\epsilon$ will be rather small, which will force $\delta$ to be, and the restriction $\delta \lt 1$ doesn't matter.  However, if your normal answer were $\delta = \epsilon$, say, a mean spirited adversary could give you $\epsilon = 2$ and have you fail.
A: Let $\epsilon>0$. So, we want to show that
$$|x-2||x+3|<\epsilon$$ whenever $0<|x-2|<\delta$ for some $\delta>0$.
Your professor said that you need to restrict one of the absolute values. He meant that we need to assume first that $|x-2|<1$. This must be the trick so that we can find a bound for the factor $|x+3|$. Now, we get $-1<x-2<1$, that is, $1<x<3$. With this, we get $4<x+3<6$. This means that $|x+3|<6$ whenever $|x-2|<1$. Then we define $$\delta=\min\big\{1,\frac{\epsilon}{6}\big\}.$$ Then $\delta\leq 1$ and $\delta\leq \frac{\epsilon}{6}$. Hence, if $0<|x-2|<\delta$ then of course $|x-2|<1$ so that $|x+3|<6$ and so
$$\begin{align}
|(x^2+x-1)-5|&=|x^2+x-6|\\
&=|x-2|\cdot|x+3|\\
&<\delta\cdot6\leq \frac{\epsilon}{6}\cdot 6=\epsilon.
\end{align}$$ Done.
A: Given $\epsilon>0$, we need $\delta>0$ such that if $|x-2|<\delta$, then $|(x^2+x-1)-5|<\epsilon$. Note that
$$
|(x^2+x-1)-5|<\epsilon\Longleftrightarrow |x^2+x-6|<\epsilon\Longleftrightarrow |x-2||x+3|<\epsilon. 
$$
Suppose we first require that $|x-2|<1$; that is, we are letting $\delta=1$ in this case. Then
$$
-1<x-2<1\Longleftrightarrow 4<x+3<6\Longleftrightarrow|x+3|<6.
$$
Hence, choose $\delta=\min\{1,\frac{\epsilon}{6}\}$. Then, if $|x-2|<\delta$, we have $|x-2|<\frac{\epsilon}{6}$, and $|x+3|<6$. Thus,
$$
|(x^2+x-1)-5|=|x^2+x-6|=|x-2||x+3|<\frac{\epsilon}{6}\cdot6=\epsilon,
$$
as desired.
A: You want to show $$ \forall \epsilon > 0, \exists \delta > 0, |x^2+x-1 -L|<\epsilon : 0<|x-2|<\delta. $$
Hence,
$$ |x^2+x-1 -L| = |x+3||x-2|.$$
Since we know $$ 0 < |x-2| < \delta $$
$$ \Rightarrow 0 < (x-2)^2 < \delta $$
Let $\delta < 1$, without loss of generality. 
Then $$ 0 < x^2 -4x +4 < 1$$
$$ \Rightarrow 1<x<3$$
$$ \Rightarrow 4<x+3<6$$
Then, $$ |x^2+x-1 -L| < 6\delta < \epsilon $$
Hence 
$$ \delta < \frac{\epsilon}{6}$$
Take $\delta:=min\{1,\frac{\epsilon}{6}\}$. Then your statement is true.
