
Limit algebra
Please explain to me how:
$\displaystyle \lim_{x\to\\infty} \frac{3x}{4x\sqrt{16x^2(1\frac{3}{16x})}}$
becomes
$\displaystyle \lim_{x\to\\infty} \frac{3x}{4x(4x)\sqrt{(1\frac{3}{16x})}}$
Bringing 16x^{2} 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?

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.)
$\displaystyle = \lim_{x\to\\infty} \frac{3x}{4x4x\sqrt{(1\frac{3}{16x})}}$ (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.)