evaluate: lim x-0 [1-cosx (cos2x)^1/2] / x^2 - Maths - Limits and Derivatives

Question

evaluate:
lim x-0 [1-cosx (cos2x)^1/2] / x^2

Santosh Kumar · Accepted Answer

$\lim_{x \to 0} (\frac{1 - \cos x \sqrt{\cos 2 x}}{x^{2}}) = \lim_{x \to 0} \frac{(1 - \cos x \sqrt{\cos 2 x})}{x^{2}} \times \frac{(1 + \cos x \sqrt{\cos 2 x})}{(1 + \cos x \sqrt{\cos 2 x})} = \lim_{x \to 0} \frac{(1)^{2} - (\cos x \sqrt{\cos 2 x})^{2}}{x^{2} (1 + \cos x \sqrt{\cos 2 x})} = \lim_{x \to 0} \frac{1 - \cos^{2} x . \cos 2 x}{x^{2} (1 + \cos x \sqrt{\cos 2 x})} = \lim_{x \to 0} \frac{1 - (\cos^{2} x) (2 \cos^{2} x - 1)}{x^{2} (1 + \cos x \sqrt{\cos 2 x})} = \lim_{x \to 0} \frac{1 - 2 \cos^{4} x + \cos^{2} x}{x^{2} (1 + \cos x \sqrt{\cos 2 x})}$

$= \lim_{x \to 0} \frac{- 2 \cos^{4} x + \cos^{2} x + 1}{x^{2} (1 + \cos x \sqrt{\cos 2 x})} = \lim_{x \to 0} \frac{- (2 \cos^{4} x - \cos^{2} x - 1)}{x^{2} (1 + \cos x \sqrt{\cos 2 x})} = \lim_{x \to 0} \frac{- (2 \cos^{4} x - 2 \cos^{2} x + \cos^{2} x - 1)}{x^{2} (1 + \cos x \sqrt{\cos 2 x})} = \lim_{x \to 0} \frac{- [2 \cos^{2} x (\cos^{2} x - 1) + 1 (\cos^{2} x - 1)]}{x^{2} (1 + \cos x \sqrt{\cos 2 x})} = \lim_{x \to 0} \frac{- (2 \cos^{2} x + 1) (\cos^{2} x - 1)}{x^{2} (1 + \cos x \sqrt{\cos 2 x})} = \lim_{x \to 0} \frac{(1 + 2 \cos^{2} x) (1 - \cos^{2} x)}{x^{2} (1 + \cos x \sqrt{\cos 2 x})}$

$= \lim_{x \to 0} \frac{(1 + 2 \cos^{2} x) (\sin^{2} x)}{x^{2} (1 + \cos x \sqrt{\cos 2 x})} = \lim_{x \to 0} (\frac{\sin^{2} x}{x^{2}}) \times \lim_{x \to 0} (\frac{1 + 2 \cos^{2} x}{1 + \cos x \sqrt{\cos 2 x}}) = 1 \times [\frac{1 + 2 \cos 0}{1 + \cos 0 (\sqrt{\cos 0})}] = 1 \times (\frac{1 + 2}{1 + 1}) = 1 \times \frac{3}{2} = \frac{3}{2}$