A) 1
B) -1
C) zero
D) does not exist
Correct Answer: D
Solution :
\[\underset{x\to 0}{\mathop{\lim }}\,\frac{\sqrt{1-\cos 2x}}{\sqrt{2}\cdot x}=\underset{x\to 0}{\mathop{\lim }}\,\frac{\sqrt{1-(1-2{{\sin }^{2}}x)}}{\sqrt{2}\cdot x}\] \[\underset{x\to 0}{\mathop{\lim }}\,\frac{\sqrt{1-\cos 2x}}{\sqrt{2}\cdot x}=\underset{x\to 0}{\mathop{\lim }}\,\frac{|\sin x|}{x}\] Let\[f(x)=\frac{|\sin x|}{x}\] Then, \[f(0+0)=\underset{h\to 0}{\mathop{\lim }}\,\frac{|\sin (0+h)|}{\sqrt{2}\cdot x}=\underset{h\to 0}{\mathop{\lim }}\,\frac{\sinh }{h}=1\] and \[f(0-0)=\underset{h\to 0}{\mathop{\lim }}\,\frac{|\sin (0-h)|}{-h}=\underset{h\to 0}{\mathop{\lim }}\,\frac{\sinh }{-h}=-1\] \[\because \] \[f(0+0)\ne f(0-0)\] \[\therefore \]The limit does not exist.
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