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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.

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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
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2 Answers

up vote 3 down vote accepted

HINT:

$$\frac{x^2+x-6}{x^2-9}-\frac56=\frac{6x^2+6x-36-(5x^2-45)}{6(x^2-9)}=\frac{x^2+6x+9}{6(x^2-9)}=\frac{(x+3)^2}{6(x-3)(x+3)}$$

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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
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Hint:

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

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