# Is this epsilon-delta proof valid?

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.

You have $x^2 + 6x + 8 = (x+2)(x + 4).$ Try factoring and canceling.
• I don't see how the first inequality follows. You have a problem that depends on the sign of $x + 4$. – ncmathsadist Jan 2 '14 at 20:10