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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

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    $\begingroup$ Try factoring $x^3-8$ $\endgroup$ – ClassicStyle Oct 6 '14 at 1:57
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    $\begingroup$ Imagine for a moment that this is like any other $\delta$-$\epsilon$ proof. Try it. If you run into trouble, you'll have a more specific question to ask. $\endgroup$ – David K Oct 6 '14 at 1:59
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    $\begingroup$ You'll have to explain why $|x + 4| \leq 6$. $\endgroup$ – layman Oct 6 '14 at 2:23
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    $\begingroup$ $|x + 4| \leq 6$ only if $x \leq 2$. We could have $x \geq 2$, though, right? For example, if we are taking the limit as $x$ approaches $2$ from the right (i.e., $\lim \limits_{x \to 2^{+}}$...) By the way, you are very close. It is just this last detail that is missing. $\endgroup$ – layman Oct 6 '14 at 2:26
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    $\begingroup$ Sure, let me write up my suggestion as a post. $\endgroup$ – layman Oct 6 '14 at 2:30
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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.

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  • $\begingroup$ Thanks for the detailed write up. Could we conclude that $\delta = \frac{\epsilon}{7}$ and still be correct? Or is it that $\delta = \min{1,\frac{\epsilon}{7}}$ is the only correct answer? $\endgroup$ – Pubbie Oct 6 '14 at 2:43
  • $\begingroup$ @Pubbie We got the $\frac{\epsilon}{7}$ by first assuming $\delta < 1$, since this assumption gave us $|x + 4| < 7$. So we need both $\delta < 1$ and $\delta < \frac{\epsilon}{7}$ to be true. The only way to ensure both of these are true in all cases is to take the minimum. $\epsilon$ was arbitrary, so if we took $\epsilon > 7$, then $\frac{\epsilon}{7}$ would be a number greater than $1$, and so $\delta = \frac{\epsilon}{7}$ would not satisfy $\delta < 1$. So we do need that $\delta = \min \{ 1, \frac{\epsilon}{7} \}$. It is the simplest way to take care of all cases at once. $\endgroup$ – layman Oct 6 '14 at 2:46
  • $\begingroup$ Ah! Thank you so much! $\endgroup$ – Pubbie Oct 6 '14 at 2:55
  • $\begingroup$ @Pubbie You're welcome! $\endgroup$ – layman Oct 6 '14 at 2:55

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