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What is the value of \(\rm \mathop {\lim} \limits_{x\rightarrow 0}\frac{(1-\cos 2x)^3}{x^6}\)= ?
1. 2
2. 4
3. 8
4. 16

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Correct Answer - Option 3 : 8

Concept:

cos 2x = 1 - 2sin2

\(\rm lim_{x\rightarrow 0}\frac{sinx}{x}=1\)

 

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.

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