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Question: \(\lim _ { x \rightarrow 0 }\) \(\frac{1 - \cos^{3}x}{x\sin x\cos x}\)is equal to...

lim⁡x→0\lim _ { x \rightarrow 0 } 1−cos⁡3xxsin⁡xcos⁡x\frac{1 - \cos^{3}x}{x\sin x\cos x}is equal to

A

2/5

B

3/5

C

3/2

D

¾

Answer

3/2

Explanation

Solution

lim⁡x→0\lim _ { x \rightarrow 0 } 1−cos⁡3xxsin⁡xcos⁡x\frac{1 - \cos^{3}x}{x\sin x\cos x}= lim⁡x→0\lim _ { x \rightarrow 0 }

(1−cos⁡x)(1+cos⁡2x+cos⁡x)xsin⁡xcos⁡x\frac{(1 - \cos x)(1 + \cos^{2}x + \cos x)}{x\sin x\cos x}

= lim⁡x→0\lim _ { x \rightarrow 0 } 2sin⁡2(x/2)(1+cos⁡2x+cos⁡x)2xsin⁡(x/2)cos⁡(x/2)cos⁡(x/2)cos⁡x\frac{2\sin^{2}(x/2)(1 + \cos^{2}x + \cos x)}{2x\sin(x/2)\cos(x/2)\cos(x/2)\cos x}

= lim⁡x→0\lim_{x \rightarrow 0} 12\frac{1}{2}. sin⁡(x/2)x/2\frac{\sin(x/2)}{x/2}. 1+cos⁡2x+cos⁡xcos⁡(x/2)cos⁡x\frac{1 + \cos^{2}x + \cos x}{\cos(x/2)\cos x}

= 12\frac{1}{2}. 1. 31\frac{3}{1}= 32\frac{3}{2}