A) \[\lambda \]
B) \[-1\]
C) zero
D) does not exist
Correct Answer: D
Solution :
Limit of a function exists only, if LHL = RHL \[\underset{x\to 0}{\mathop{\lim }}\,\frac{\sqrt{1-\cos 2x}}{\sqrt{2}x}=\underset{x\to 0}{\mathop{\lim }}\,\frac{\sqrt{2.{{\sin }^{2}}x}}{\sqrt{2}x}\] \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{\sqrt{2}\left| \sin x \right|}{\sqrt{2}x}=\underset{x\to 0}{\mathop{\lim }}\,\frac{\left| \sin x \right|}{x}\] Let \[f(x)=\frac{\left| \sin x \right|}{x}\] Now, LHL= \[\underset{h\to 0}{\mathop{\lim }}\,\frac{\left| \sin (0-h) \right|}{0-h}\] \[=\underset{h\to 0}{\mathop{\lim }}\,\frac{\sin \,h}{-h}=-1\] and RHL = \[=\underset{h\to 0}{\mathop{\lim }}\,\frac{\left| \sin \,(0+h) \right|}{0+h}\] \[=\underset{h\to 0}{\mathop{\lim }}\,\frac{\sin \,h}{-h}=1\] \[\Rightarrow \] LHL \[\ne \] RHL \[\therefore \,\,\underset{x\to 0}{\mathop{\lim }}\,\frac{\left| \sin x \right|}{x}\] does not exist.
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