Proving a limit using epsilon delta definition I'm trying to prove a limit (by showing that I can find a delta for all epsilon) using the $\epsilon$, $\delta$ definition but I'm stuck.
$$\lim_{x\to2}\left(x^2+2x-7\right)\ = 1$$
So I got to this point where I factored the polynomial and separated the absolute values but I don't know what to do next.
$$|x^2+2x-7-1| < \epsilon \Rightarrow |x-2| \lt \delta$$
$$|x+4||x-2| < \epsilon \Rightarrow |x-2| < \delta$$
Can someone help nudge me in the right direction?
 A: $$\lim_{x\to2}\left(x^2+2x-7\right)\ = 1$$
For every $\epsilon > 0$, there exists a $\delta >0$ such that
$|x-2| < \delta \implies |(x^2+2x-7) - 1| < \epsilon$.
Very often, you solve these problems by looking at what $\epsilon$ needs to do and then working backwards to what $\delta$ needs to do. In this case
$$|(x^2+2x-7) - 1| = |x^2+2x-8| = |(x+4)(x-2)| = |x+4|\,|x-2|$$
So, we need to make $|x+4|\,|x-2| < \epsilon$. We know we are going to make $|x-2| < \delta$, but what do we do with $|x+4|$?
\begin{align}
   |x-2| < \delta
   &\implies 2-\delta < x < 2 + \delta \\
   &\implies 4-\delta < x + 4 < 4 + \delta
\end{align}
The trick is to limit the size of $\delta$. There is no fixed limit that you need to use. Just pick one. I think $10$ is a nice round number so I am going to say, suppose $0 < \delta < 2$. Then
\begin{align}
   |x-2| < \delta  \; \text{and} \; (0 < \delta < 2)
   &\implies (2-\delta < x < 2 + \delta) \; \text{and} \; (0<\delta<2) \\
   &\implies (6-\delta < x+4 < 6+\delta) \; \text{and} \; (-2<-\delta<0)\\
   &\implies 4 < x + 4 < 8 \\
   &\implies |x+4| < 10 \\
   &\implies |x+4||x-2| < 10\delta \\
\end{align}
You should see that we now solve $10\delta < \epsilon$ for $\delta$. We get $\delta < \dfrac{\epsilon}{10}$. But wait! We made an assumption that $\delta < 2$. That's very easy to fix. Our final formula is
$\delta = \min\left\{2, \dfrac{\epsilon}{10} \right\}$.
Then we get our proof by adding one more line to the previous argument.
\begin{align}
   |x-2| < \delta  \; \text{and} \; (0 < \delta < 2)
   &\implies (2-\delta < x < 2 + \delta) \; \text{and} \; (0 < \delta < 2) \\
   &\implies (6-\delta < x+4 < 6+\delta) \; \text{and} \; (0<\delta<2) \\
   &\implies 4 < x + 4 < 8 \\
   &\implies |x+4| < 10 \\
   &\implies |x+4||x-2| < 10\delta \\
   &\implies |(x^2+2x-7) - 1| < \epsilon
\end{align}
A: Suppose $|x-2| < \epsilon$. We can then write:
$$|x^2 + 2x -7 -1| = |x+4| |x-2| \leq \epsilon (6+\epsilon),$$where the last term comes from the fact that $2-\epsilon < x<2+\epsilon$. Now, choose $\delta = \epsilon (6+\epsilon)$. 
A: I dislike the way examples of this sort are usually presented. Here's how I would do it.
Let $f(x) = x^2+2x-7$. Let $\epsilon>0$. We need to find $\delta$ so
that $$| f(2+h) - 1 | < \epsilon $$ whenever $0<|h|<\delta$. To that end, let's estimate $| f(2+h) - 1 |$.
$$|f(2+h)-1| = |6h + h^2  | \leq 6|h| + |h|^2.$$
The intuition now is that if $|h|$ is small, then $|h|^2$ is even smaller,
so that the $|h|^2$ can essentially be ignored. To make this precise, we note that if $|h|<1$, then $|h|^2< |h|$, so
$$|f(2+h)-1| < 7|h|.$$
So to get $|f(2+h)-1|<\epsilon$, it suffices to have $|h|<1$ and $7|h|<\epsilon$,
or in other words, $|h|<\min(1, \frac{\epsilon}{7})$.
Thus, $$\delta = \min(1, \frac{\epsilon}{7})$$ satisfies our requirements.
This approach emphasizes the concept of bounding quantities. It also suggests that we should look at orders of growth of different quantities. (In this case $|h|^2$ is of smaller order than $|h|$ near $0$.) 
