Compare the Limits

Use L'Hôpital's Rule to solve this problem. $a = \lim_{x \to 0}\frac{\ln[\cos 2x]}{3x^{2}} = \lim_{x \to 0}\frac{\frac{-2\sin 2x}{\cos 2x}}{6x} = \lim_{x \to 0}\frac{-2\tan 2x}{6x} = \lim_{x \rightarrow 0} \frac{-4\sec^{2}2x}{6} = \frac{-4\sec^{2}0}{6} = \boxed{-\frac{2}{3}}$ $b = \lim_{x \rightarrow 0} \frac{\sin^{2}2x}{x(1 - e^{x})} = \lim_{x \rightarrow 0} \frac{\frac{1-\cos4x}{2}}{x-xe^{x}} = \lim_{x \rightarrow 0} \frac{2\sin4x}{1 - xe^{x} - e^{x}} = \lim_{x \rightarrow 0} \frac{-8e^{-x}\cos4x}{x+2} = \frac{-8e^{0}\cos0}{2} = \boxed{-4}$ $c = \lim_{x \rightarrow 1} \frac{\sqrt{x}-x}{\ln x} = \lim_{x \rightarrow 1} \frac{\frac{1}{2\sqrt{x}}-1}{\frac{1}{x}} = \lim_{x \rightarrow 1} (\frac{\sqrt{x}}{2} - \frac{1}{x}) = \frac{\sqrt{1}}{2} - \frac{1}{1} = \boxed{-\frac{1}{2}}$ $-\frac{1}{2} > -\frac{2}{3} > -4$ $\therefore \boxed{c > a > b}$