1. ## Limit algebra

$\displaystyle \lim_{x\to\--\infty} \frac{3x}{4x-\sqrt{16x^2(1-\frac{3}{16x})}}$
becomes
$\displaystyle \lim_{x\to\--\infty} \frac{3x}{4x-(-4x)\sqrt{(1-\frac{3}{16x})}}$

Bringing 16x2 out from the radical becomes negative because x is approaching negative infinity? I don't quite understand why the other coefficients would not automatically become negative, or more importantly why is this one an exception??. Is there some vast concept at work here, or should I just devote this "rule" to route memorization?

2. ## Re: Limit algebra

$\displaystyle \lim_{x\to\--\infty} \frac{3x}{4x-\sqrt{16x^2(1-\frac{3}{16x})}}$

$\displaystyle = \lim_{x\to\--\infty} \frac{3x}{4x-\sqrt{16x^2}\sqrt{(1-\frac{3}{16x})}}$ (Note both factors under the original square root are positive.)

$\displaystyle = \lim_{x\to\--\infty} \frac{3x}{4x-|4x|\sqrt{(1-\frac{3}{16x})}}$ (Note that square root sign always means the POSITIVE square root.)

$\displaystyle \lim_{x\to\--\infty} \frac{3x}{4x-(-4x)\sqrt{(1-\frac{3}{16x})}}$ (Note that x negative, so |x| = -x.)