A) 0
B) 1
C) \[\infty \]
D) None of these
Correct Answer: A
Solution :
\[\underset{x\to 0}{\mathop{\lim }}\,\,\,\frac{\log \cos x}{x}=\underset{x\to 0}{\mathop{\lim }}\,\,\,\frac{\log \,\left[ 1-2{{\sin }^{2}}\frac{x}{2} \right]}{x}\] \[=\underset{x\to 0}{\mathop{\lim }}\,\,\,\frac{-\,\left[ 2\,{{\sin }^{2}}\frac{x}{2}+{{\left( \frac{2\,{{\sin }^{2}}\frac{x}{2}}{2} \right)}^{2}}+...... \right]}{x}=0\] Aliter : Apply L-Hospital?s rule, \[\underset{x\to 0}{\mathop{\lim }}\,\,\frac{\log \cos x}{x}=\underset{x\to 0}{\mathop{\lim }}\,\,\frac{-\tan x}{1}=0.\]
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