limx→01−cos x cos 2x cos 3xsin22x is equal to
74
∵cos x cos 2x cos 3x
= 12{(2cos x cos 2x)cos 3x}
= 12{(cos 3x+cos x)cos 3x}
= 14{(2cos2 3x+2cos 3x cos 3x)}
= 14{1+cos 6x+cos 4x+cos 2x}
With this our expression becomes -
limx→01−14{1+cos 6x+cos 4x+cos 2x}sin22x
Now we can apply L'Hospital rule -
limx→0−14{−6sin 6x−4sin 4x−2sin 2x}4.sin 2x.cos 2x
limx→0−14{−6sin 6x−4sin 4x−2sin 2x}2.sin 4x
With the help of standard limits we'll get the above expression equal to 7/4.