A) \[-\frac{1}{\sqrt{2}}\]
B) \[\frac{1}{\sqrt{2}}\]
C) \[0\]
D) Does not exist
Correct Answer: B
Solution :
\[\sqrt{1-\cos x}=\left\{ \begin{matrix} -\sqrt{2}\sin \frac{x}{2}, & x<0 \\ \sqrt{2}\sin \frac{x}{2}, & x\ge 0 \\ \end{matrix} \right.\] \[\therefore \] \[\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\frac{\sqrt{1-\cos x}}{x}=\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,-\frac{\sqrt{2}\sin \frac{x}{2}}{x}=-\frac{1}{\sqrt{2}}\] and \[\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\frac{\sqrt{1-\cos x}}{x}=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\frac{\sqrt{2}\sin \frac{x}{2}}{x}=\frac{1}{\sqrt{2}}\] \[\because \] \[\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\frac{\sqrt{1-\cos x}}{x}\ne \underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\frac{\sqrt{1-\cos x}}{x}\] So, \[\underset{x\to 0}{\mathop{\lim }}\,\frac{\sqrt{1-\cos x}}{x}\]does not exist.
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