A) 0
B) 1
C) \[\frac{1}{2}\]
D) \[-\frac{1}{2}\]
Correct Answer: D
Solution :
Applying L-Hospital?s rule, \[\underset{x\to 0}{\mathop{\lim }}\,\frac{\ln (\cos x)}{{{x}^{2}}}\]\[=\underset{x\to 0}{\mathop{\lim }}\,\frac{-\tan x}{2x}\]\[=\underset{x\to 0}{\mathop{\lim }}\,\frac{-{{\sec }^{2}}x}{2}=\frac{-1}{2}\].
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