Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
You could look at… which is almost the same question. – Ross Millikan Jan 24 '13 at 19:18

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

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.