Question

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

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

Answer 1

The correct option is C

$\frac{3}{2}$

Explanation for the correct option:

Finding the value of the given limit:

$$\begin{gathered} \underset{x\to 0}{\lim }\frac{[1-{\cos }^{3}x]}{x\sin x\cos x} \\ =\underset{x\to 0}{\lim }\frac{(1-\cos x)(1+{\cos }^{2}x+\cos x)}{x\sin x\cos x}\left[\because a^3-b^3=(a-b)(a^2+b^2-ab)\right] \\ =\underset{x\to 0}{\lim }\frac{1-\cos \left(x\right)}{x^2}\times \frac{x}{\sin \left(x\right)}\times \frac{1+{\cos }^{2}x+\cos x}{\cos x} \end{gathered}$$

Applying the limits,

$$\begin{gathered} =\frac{1}{2}\times \left(1\right)\times \left(\frac{1+1+1}{1}\right) \\ =\frac{1}{2}\times \left(\frac{3}{1}\right) \\ =\frac{3}{2} \end{gathered}$$

Therefore, option (C) is correct.