Question

Write the value of $\underset{x\to 0}{\lim }\frac{\sqrt{1-\cos 2x}}{x}.$

Answer 1

$$\begin{gathered} \underset{x\to 0}{\lim }\left(\frac{\sqrt{1-\cos 2x}}{x}\right) \\ =\underset{x\to 0}{\lim }\left(\frac{\sqrt{2{\sin }^{2}x}}{x}\right) \\ =\sqrt{2}\underset{x\to 0}{\lim }\left(\frac{\sqrt{{\sin }^{2}x}}{x}\right) \\ =\sqrt{2}\underset{x\to 0}{\lim }\left(\frac{\left|\sin x\right|}{x}\right) \\ \mathrm{LHL}: \\ =\sqrt{2}\underset{x\to 0^{-}}{\lim }\left(\frac{\left|\sin x\right|}{x}\right) \\ \mathrm{Let}x=0-h,\mathrm{where}h\to 0. \\ =\sqrt{2}\underset{h\to 0}{\lim }\left(\frac{\left|\sin \left(-h\right)\right|}{-h}\right) \\ =\sqrt{2}\underset{h\to 0}{\lim }\left(\frac{\sin h}{-h}\right) \\ =-\sqrt{2} \\ \mathrm{RHL}: \\ =\sqrt{2}\underset{x\to 0^{+}}{\lim }\left(\frac{\left|\sin x\right|}{x}\right) \\ \mathrm{Let}x=0+h,\mathrm{where}h\to 0. \\ =\sqrt{2}\underset{h\to 0}{\lim }\frac{\left|\sin h\right|}{h}=\sqrt{2} \\ \mathrm{LHL}\ne \mathrm{RHL} \\ \mathrm{Thus},\underset{x\to 0}{\lim }\left(\frac{\sqrt{1-\cos 2x}}{x}\right)\mathrm{does}\mathrm{not}\mathrm{exist}. \end{gathered}$$