Correct Answer - Option 3 : 8
Concept:
cos 2x = 1 - 2sin2x
\(\rm lim_{x\rightarrow 0}\frac{sinx}{x}=1\)
Calculation:
Given: \(\rm \mathop {\lim} \limits_{x\rightarrow 0}\frac{(1-\cos 2x)^3}{x^6}\)
We know, cos 2x = 1 - 2sin2x ⇒ 1 - cos 2x = 2sin2x
\(\rm \mathop {\lim} \limits_{x\rightarrow 0}\frac{(1-\cos 2x)^3}{x^6}\)
= \(\rm \mathop {\lim} \limits_{x\rightarrow 0}\frac{(2sin^2x)^3}{x^6}\)
= \(\rm \mathop {\lim} \limits_{x\rightarrow 0}\frac{8sin^6x}{x^6}\)
= \(\rm 8\times \mathop {\lim} \limits_{x\rightarrow 0} (\frac{sinx}{x})^6\)
= 8 (∵ \(\rm\mathop {\lim} \limits_{x\rightarrow 0}\frac{sinx}{x}=1\) )
Hence, option (3) is correct.