Question

$\underset{x\to 0}{lim}\frac{1-{\cos }^{3}x}{x\sin 2x}$=

• A. $\frac{1}{2}$
• B. $\frac{3}{2}$
• C. $\frac{3}{4}$
• D. $\frac{1}{4}$

Answer 1

Answer: (C)

$\frac{3}{4}$

$\underset{x\to 0}{lim}\frac{1-{\cos }^{3}x}{x\sin 2x}=\underset{x\to 0}{lim}\frac{(1-\cos x)(1+\cos x+{\cos }^{2}x)}{x\sin 2x}$

$=\underset{x\to 0}{lim}\frac{2{\sin }^2\frac{x}{2}(1+\cos x+{\cos }^{2}x)}{4x\cos x\sin \frac{x}{2}\cos \frac{x}{2}}=\underset{x\to 0}{lim}\frac{\sin \frac{x}{2}(1+\cos x+{\cos }^{2}x)}{4\times \frac{x}{2}\cos \frac{x}{2}\times \cos x}=\frac{3}{4}$...... As$\{{lim}_{\frac{x}{2}\to 0}\frac{\sin \frac{x}{2}}{\frac{x}{2}}=0\}$