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Prove using $\epsilon- \delta$ that $ \displaystyle \lim_{x \to -4} \frac { x^2 + 6x + 8 }{x + 4} = - 2 $. Here's a proposed proof:


For $\delta \leq 1$, i.e. $ | x + 4 | < 1 $ which guarantees $x < -1 $, one can argue:

$ \left| \dfrac { x^2 + 6x + 8 }{x + 4} + 2 \right| = \left| \dfrac { x^2 + 8x + 16}{x + 4}\right| < \left| \dfrac { x^2 + 8x + 16}{x}\right| < |x^2 + 8x + 16| = |(x+4)^2| = (x+4)^2 \ . $

Let's require $(x+4)^2 < \epsilon$, which implies $ | x + 4 | < \sqrt \epsilon $. Therefore we have $\delta = \min \{1, \sqrt \epsilon \}$.


Is it a valid proof or are there any loopholes I'm unaware of? Side-note: I realize there are different -- and perhaps simpler -- ways to prove this, I just want to see if this very approach is valid.

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You have $x^2 + 6x + 8 = (x+2)(x + 4).$ Try factoring and canceling.

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  • $\begingroup$ Yes I am aware of that. I just would like to know whether the proposed proof is valid. Thank you anyway. $\endgroup$ – kingemailguest Jan 2 '14 at 20:08
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    $\begingroup$ I don't see how the first inequality follows. You have a problem that depends on the sign of $x + 4$. $\endgroup$ – ncmathsadist Jan 2 '14 at 20:10
  • $\begingroup$ Oh! You are right, it does not follow. Seems there was a loophole after all. Thank you! $\endgroup$ – kingemailguest Jan 2 '14 at 20:24

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