Delta Epsilon Quotient Proof I'm trying to justify $\lim_{x\rightarrow x_0} \frac{x^3-8}{x-2} = 12$, where $x_0=2$ using $\delta-\epsilon$ proofs. I've never done one of these proofs where $\lim_{x\rightarrow x_0} \frac{L_1}{L_2} = \frac{0}{0}$. 
Thank!
Edit: Came up with a proof. Could someone verify it? 
Given $\epsilon >0$ let $\delta = \epsilon$ \ $6$. This implies that $|x-x_0|<\delta \implies |x-2|<\frac{\epsilon}{6}$ $$\implies |x-2||x+4|<6\frac{\epsilon}{6}$$ since $|x+4|\le 6$. Therefore, we have that $|x^2+2x-8|<\epsilon \implies |(x^2+2x+4)-12|<\epsilon$, which verifies that $\lim_{x\rightarrow 2 } \frac{x^3-8}{x-2} = 12$. QED
 A: Your proof is a huge step in the right direction.  But you can't conclude that $|x + 4| \leq 6$ because, as we said, we could approach $2$ from the right.
So what can we do?  Well, you might not like this (I didn't when I first saw it but now it makes perfect sense), but we can assume without loss of generality that $\delta < 1$.  Why?
Think about it.  Given $\epsilon > 0$, we want to find $\delta > 0$ such that $|x - 2| < \delta$ implies $|f(x) - 12| < \epsilon$, right?  But as soon as we have found one $\delta > 0$, we have found infinitely many because any positive $\gamma < \delta$ also works.  Specifically, after finding $\delta > 0$, we are assured $|x - 2| < \delta$ implies $|f(x) - 12| < \epsilon$.  But since $\gamma < \delta$, then $|x - 2| < \gamma$ implies $|f(x) - 12| < \epsilon$ (since we get $|x - 2| < \gamma < \delta$).
Anyway, the point of what I wrote above is that when we find one $\delta$ that works, any positive number smaller than $\delta$ will also work.  Because of this, we can assume without loss of generality that $\delta < 1$ (if it doesn't work for $\delta < 1$, then it won't work for $\delta \geq 1$ either because of what we just said, that if it works for a larger number, it should work for all smaller numbers -- so if it worked for a number larger than or equal to $1$, it should work for numbers $< 1$...).
So, assuming $\delta < 1$ (you could replace $1$ with any other positive number and still get the same proof, although with a slightly different $\delta$), we have $|x^{2} + 2x - 8| = |x + 4||x - 2|$.  Well, we know $|x - 2| < \delta < 1$, but we also want to bound $|x + 4|$.  We can do that because we have $|x - 2| < 1$, as shown below:
$|x - 2| < 1 \implies -1 < x - 2 < 1 \implies 1 < x < 3$, so we get that $x < 3$, and consequently, $|x + 4| < |3 + 4| = 7$.
Finally, we assumed WLOG that $\delta < 1$.  But now we want $\delta < \frac{\epsilon}{7}$ since this would mean $|x + 4||x - 2| < 7*\frac{\epsilon}{7} = \epsilon$.
Since we want $\delta < 1$ AND $\delta < \frac{\epsilon}{7}$, let's choose $\delta = \min \{ 1, \frac{\epsilon}{7}   \}$.  Then we get $|x + 4||x - 2| < 7 * \frac{\epsilon}{7} = \epsilon$ if $|x - 2| < \delta$, as desired.
