Question

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

• A.
• B.
• C.
• D.

Answer 1

The correct option is C

$\because cosxcos2xcos3x=\frac{1}{2}\left\{\right(2cosxcos2x\left)cos3x\right\}=\frac{1}{2}\left\{\right(cos3x+cosx\left)cos3x\right\}=\frac{1}{4}\{2co{s}^{2}3x+2cos3xcosx\}=\frac{1}{4}\{1+cos6x+cos4x+cos2x\}\therefore 1-cosxcox2xcos3x=1-\frac{1}{4}\{1+cos6x+cos4x+cos2x\}=\frac{1}{4}\{4-1-cos6x-cos4x-cos2x\}=\frac{1}{4}\left\{\right(1-cos6x)+(1-cos4x)+(1-cos2x\left)\right\}=\frac{1}{4}\{2si{n}^{2}3x+2si{n}^{2}2x+2si{n}^{2}x\}=\frac{1}{2}(si{n}^{2}3x+si{n}^{2}2x+si{n}^{2}x)\therefore \underset{x\to 0}{lim}\frac{1-cosxcos2xcos3x}{si{n}^{2}2x}=\frac{1}{2}\underset{x\to 0}{lim}\left(\frac{si{n}^{2}3x+si{n}^{2}2x+si{n}^{2}x}{si{n}^{2}2x}\right)=\frac{1}{2}\underset{x\to 0}{lim}\left\{\frac{9.\frac{si{n}^{2}3x}{9x^2}+4.\frac{si{n}^{2}2x}{4x^2}+\frac{si{n}^{2}x}{x^2}}{4.\frac{si{n}^{2}2x}{4x^2}}\right\}=\frac{1}{2}\left\{\frac{9.1+4.1+1.1}{4}\right\}=\frac{7}{4}$