# 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?

$$\lim_{x\to2}\left(x^2+2x-7\right)\ = 1$$

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

$$|(x^2+2x-7) - 1| = |x^2+2x-8| = |(x+6)(x-2)| = |x+6|\,|x-2|$$

So, we need to make $|x+6|\,|x-2| < \epsilon$. We know we are going to make $|x-2| < \delta$, but what do we do with $|x+6|$?

\begin{align} |x-2| < \delta &\implies 2-\delta < x < 2 + \delta \\ &\implies 8-\delta < x + 6 < 8 + \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 (8-\delta < x+6 < 8+\delta) \; \text{and} \; (0<\delta<2) \\ &\implies 6 < x + 6 < 10 \\ &\implies |x+6| < 10 \\ &\implies |x+6||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 (8-\delta < x+6 < 8+\delta) \; \text{and} \; (0<\delta<2) \\ &\implies 6 < x + 6 < 10 \\ &\implies |x+6| < 10 \\ &\implies |x+6||x-2| < 10\delta \\ &\implies |(x^2+2x-7) - 1| < \epsilon \end{align}

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 $\delta = \epsilon (6+\epsilon)$.

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 $$0$$.)