Evaluate: limit as x approaches 0 of (cos(4x)-1)/(xtan(2x))

Expression: $\lim _{x\to 0}(\frac{\cos(4x)-1}{x\tan(2x)})$

$=\lim _{x\to 0}(\frac{-4\sin(4x)}{\tan(2x)+2x\sec^{2}(2x)})$

$=\lim _{x\to 0}(\frac{-16\cos(4x)}{4\sec^{2}(2x)+8x\sec^{2}(2x)\tan(2x)})$

$=\frac{-16\cos(4\cdot 0)}{4\sec^{2}(2\cdot 0)+8\cdot 0\cdot \sec^{2}(2\cdot 0)\tan(2\cdot 0)}$

$=-4$