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 having a bit of trouble with epsilon-delta proofs of limits. I have to prove the existence of the limit $$\lim_{x \to -3} \frac{x^2 + x - 6}{x^2 - 9} = \frac{5}{6}.$$

I want to try to relate $\delta$ and $\varepsilon$ through $$0 < |x + 3| < \delta$$ and $$\left| \frac{x^2+x-6}{x^2-9} - 5/6 \right| < \varepsilon,$$ but that's where I'm stuck. Everything I've done up to this point has come out very cleanly when I try to relate $\delta$ and $\varepsilon$ (e.g. $\delta = \varepsilon/3$), but I don't see a way to do that this time.


share|cite|improve this question
I don't see why you need to use $\delta$ and $\varepsilon$ tricks here. You have something like $\displaystyle\lim_{x\rightarrow x_0}\frac{P_1(x)}{P_2(x)}$ where $P_1(x_0)=P_2(x_0)=0$. All you have to do is notice that $x_0$ is a root of $P_1$ and $P_2$, and factor those two polynoms by $x-x_0$: $P_1(x) = Q_1(x)(x-x_0)$, $P_2(x)=Q_2(x)(x-x_0)$. – S4M Sep 17 '12 at 21:10
up vote 3 down vote accepted



share|cite|improve this answer
also, perhaps you want to impose some restriction on $\delta$ such that $|x-3|$ is not too small. Usually the trick is $\delta \leq 1$ imposed by the minimum function. – James S. Cook Sep 17 '12 at 2:50
@Brian M. Scott Thank you so much! That was exactly what I needed. I just worked through the whole thing twice and everything came out. You have my eternal gratitude! – Jackson Sep 17 '12 at 3:34


$$\frac{x^2+x-6}{x^2-9}=\frac{(x+3)(x-2)}{(x+3)(x-3)}=\frac{x-2}{x-3}\,\,,\,\,x\neq \pm\,3$$

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.