Question

$\underset{x\to 0}{lim}\frac{1-cosxcos2xcos3x}{si{n}^{2}2x}$ is equal to

• A. $\frac{7}{2}$
• B. $\frac{7}{3}$
• C. $\frac{7}{4}$
• D. $\frac{7}{5}$

Answer 1

The correct option is C

$\frac{7}{4}$

$\because cosxcos2xcos3x$

= $\frac{1}{2}\left\{\right(2cosxcos2x\left)cos3x\right\}$

= $\frac{1}{2}\left\{\right(cos3x+cosx\left)cos3x\right\}$

= $\frac{1}{4}\left\{\right(2co{s}^{2}3x+2cos3xcos3x\left)\right\}$

= $\frac{1}{4}\{1+cos6x+cos4x+cos2x\}$

With this our expression becomes -
$\underset{x\to 0}{lim}\frac{1-\frac{1}{4}\{1+cos6x+cos4x+cos2x\}}{si{n}^{2}2x}$
Now we can apply L'Hospital rule -
$\underset{x\to 0}{lim}\frac{-\frac{1}{4}\{-6sin6x-4sin4x-2sin2x\}}{4.sin2x.cos2x}$
$\underset{x\to 0}{lim}\frac{-\frac{1}{4}\{-6sin6x-4sin4x-2sin2x\}}{2.sin4x}$
With the help of standard limits we'll get the above expression equal to 7/4.