A) 0
B) \[-1\]
C) ¥
D) 1
Correct Answer: D
Solution :
[d] \[\because y=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,{{\log }_{\tan x}}(\sin x)\] \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{{{\log }_{e}}\sin x}{{{\log }_{e}}\tan x}\] which is \[\frac{0}{0}\]form \[y=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\frac{\frac{1}{\sin }.\cos x}{\frac{1}{\tan x}.{{\sec }^{2}}x}\] [by L Hospital Rule] \[y=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\frac{\frac{\cos x}{\sin x}}{\frac{{{\sec }^{2}}x}{\sin x}\times \cos x}=\underset{x\to 0}{\mathop{\lim }}\,\frac{\cos x}{\sec x}\] \[=\underset{x\to 0}{\mathop{\lim }}\,{{\cos }^{2}}x=1\]
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