# Help with epsilon-delta proof of $\lim_{x \to -3} \frac{x^2 + x - 6}{x^2 - 9}$

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.

Thanks.

• 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, 2012 at 21:10

$$\frac{x^2+x-6}{x^2-9}=\frac{(x+3)(x-2)}{(x+3)(x-3)}=\frac{x-2}{x-3}\,\,,\,\,x\neq \pm\,3$$
$$\frac{x^2+x-6}{x^2-9}-\frac56=\frac{6x^2+6x-36-(5x^2-45)}{6(x^2-9)}=\frac{x^2+6x+9}{6(x^2-9)}=\frac{(x+3)^2}{6(x-3)(x+3)}$$
• 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. Sep 17, 2012 at 2:50