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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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You could look at math.stackexchange.com/questions/285980/… which is almost the same question. –  Ross Millikan Jan 24 '13 at 19:18
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1 Answer 1

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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