A) 0
B) \[\frac{1}{4}\]
C) \[\frac{1}{2}\]
D) 1
Correct Answer: C
Solution :
\[\underset{x\to 0}{\mathop{\lim }}\,\frac{1-\cos x}{{{x}^{2}}}\] \[\left( \frac{0}{0}form \right)\] \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{sinx}{2x}\] \[=\frac{1}{2}\underset{x\to 0}{\mathop{\lim }}\,\frac{sinx}{x}\] \[=\frac{1}{2}\times 1=\frac{1}{2}\]
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