A) \[1/3\]
B) \[1/2\]
C) \[1\]
D) \[2\]
Correct Answer: B
Solution :
[b] Let \[1\tan x=\frac{1}{t}\] \[\therefore \] Given limit \[=\underset{t\to {{0}^{+}}}{\mathop{\lim }}\,\frac{1}{t}\left( \frac{1}{t}.{{\log }_{e}}\left( \frac{1}{1+t} \right)+1 \right)\] \[=\underset{t\to {{0}^{+}}}{\mathop{\lim }}\,\frac{t-{{\log }_{e}}(1+t)}{{{t}^{2}}}\] \[=\underset{t\to {{0}^{+}}}{\mathop{\lim }}\,\frac{1-\frac{1}{1+t}}{2t}=\frac{1}{2}\]
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