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Find:\( \lim _{x \rightarrow 0} \frac{1-\cos x \cdot \sqrt{\cos 2 x}}{x^{2}} \) |
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Answer» \(\lim\limits_{x\to 0}\frac{1-cosx\sqrt{cos2x}}{x^2}\) (0/0 case) \(=\lim\limits_{x\to 0}\frac{-cosx\times\frac1{2\sqrt{cos2x}}\times-2sin2x+sin x\sqrt{cos x}}{2x}\) (By using D.L.H. Rule) \(=\lim\limits_{x\to 0}\frac{cos x}{\sqrt{cos2x}}\times\frac{sin 2x}{2x}+\lim\limits_{x\to 0}\frac12\frac{sin x}x\sqrt{cos2x}\) \(=\lim\limits_{x\to 0}\frac{cos x}{\sqrt{cos2x}}\) + \(\frac12\lim\limits_{x\to 0}\sqrt{cos2x}\) (\(\because\lim\limits_{x\to 0}\frac{sin x}x=1\)) = \(\frac{cos 0}{\sqrt{cos 0}}+\frac12\sqrt{cos 0}\) = 1 + 1/2 = 3/2 Hence, \(\lim\limits_{x\to 0}\frac{1-cosx\sqrt{cos2x}}{x^2}=\frac32\) |
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