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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?

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

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}

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

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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 $0$.)

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