Question

\[\lim_{x \to 0} \frac{\sqrt{1 - \cos 2x}}{x} .\]

Answer 1

\[\lim_{x \to 0} \left( \frac{\sqrt{1 - \cos 2x}}{x} \right)\]
\[ = \lim_{x \to 0} \left( \frac{\sqrt{2 \sin^2 x}}{x} \right)\]
\[ = \sqrt{2} \lim_{x \to 0} \left( \frac{\sqrt{\sin^2 x}}{x} \right)\]
\[ = \sqrt{2} \lim_{x \to 0} \left( \frac{\left| \sin x \right|}{x} \right)\]
\[\text{ LHL }: \]
\[ = \sqrt{2} \lim_{x \to 0^-} \left( \frac{\left| \sin x \right|}{x} \right)\]
\[\text{ Let x } = 0 - h, \text{ where } h \to 0 . \]
\[ = \sqrt{2} \lim_{h \to 0} \left( \frac{\left| \sin \left( - h \right) \right|}{- h} \right)\]
\[ = \sqrt{2} \lim_{h \to 0} \left( \frac{\sin h}{- h} \right)\]
\[ = - \sqrt{2}\]
\[\text{ RHL }: \]
\[ = \sqrt{2} \lim_{x \to 0^+} \left( \frac{\left| \sin x \right|}{x} \right)\]
\[\text{ Let } x = 0 + h, \text{ where } h \to 0 . \]
\[ = \sqrt{2} \lim_{h \to 0} \frac{\left| \sin h \right|}{h} = \sqrt{2}\]
\[\text{ LHL } \neq \text{ RHL }\]
\[\text{ Thus }, \lim_{x \to 0} \left( \frac{\sqrt{1 - \cos 2x}}{x} \right) \text{ does not exist } .\]