Same general strategy as with your other question: rewrite the fraction to include factors of [tex]\dfrac{\sin ax}{ax}[/tex] or [tex]\dfrac{ax}{\sin ax}[/tex].
[tex]\dfrac{\sin x\cos2x\cos x}{\sec4x\tan3x}=\dfrac{\sin x\cos x\cos2x\cos3x\cos4x}{\sin3x}[/tex]
[tex]\implies\displaystyle\lim_{x\to0}\cdots=\frac13\lim_{x\to0}\frac{\sin x}x\lim_{x\to0}\frac{3x}{\sin3x}\lim_{x\to0}\cos x\cos2x\cos3x\cos4x[/tex]
All limits approach 1, so the ultimate limit is [tex]\dfrac13[/tex].
