A) \[\frac{1}{2}\]
B) \[2\]
C) \[\sqrt{2}\]
D) None of these
Correct Answer: C
Solution :
We have \[\underset{x\to 0}{\mathop{\lim }}\,\,\frac{\sqrt{1-\cos {{x}^{2}}}}{1-\cos x}=\underset{x\to 0}{\mathop{\lim }}\,\,\,\frac{\sqrt{2\,{{\sin }^{2}}({{x}^{2}}/2)}}{2\,{{\sin }^{2}}(x/2)}\] \[=\frac{1}{\sqrt{2}}\,\underset{x\to 0}{\mathop{\lim }}\,\,\left( \frac{\frac{\sin \,({{x}^{2}}/2)}{{{x}^{2}}/2}}{{{\left( \frac{\sin \,(x/2)}{x/2} \right)}^{2}}} \right).\frac{{{x}^{2}}/2}{{{x}^{2}}/4}=\sqrt{2}\].
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