A) \[\lambda \]
B) \[-1\]
C) zero
D) does not exist
Correct Answer: D
Solution :
\[\underset{x\to 0}{\mathop{\lim }}\,\frac{\sqrt{1-\cos 2x}}{\sqrt{2}x},\] \[\Rightarrow \] \[\underset{x\to 0}{\mathop{\lim }}\,\frac{\sqrt{1-(1-2{{\sin }^{2}}x)}}{\sqrt{2}x}\] \[\underset{x\to 0}{\mathop{\lim }}\,\frac{\sqrt{2{{\sin }^{2}}x}}{\sqrt{2}x}\Rightarrow \underset{x\to 0}{\mathop{\lim }}\,\frac{|\sin x|}{x}\] So, \[f(x)=\frac{|\sin x|}{x}\] Now, \[f(0+0)=\underset{h\to 0}{\mathop{\lim }}\,\frac{|\sin (0+h)|}{0+h}\] \[=\underset{h\to 0}{\mathop{\lim }}\,\frac{\sin \,h}{h}=1\] \[f(0-0)=\underset{h\to 0}{\mathop{\lim }}\,\frac{|\sin (0-h)|}{-h}\] \[\underset{h\to 0}{\mathop{\lim }}\,\frac{\sin h}{-h}=-1\] \[\because \] \[F(0+0)\ne f(0-0)\]the limit of function does not exist.
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