A) 1
B) -1
C) 0
D) None of these
Correct Answer: C
Solution :
[c] \[\underset{x\to 0}{\mathop{\lim }}\,{{\log }_{e}}{{(sinx)}^{\tan x}}=\underset{x\to 0}{\mathop{\lim }}\,\tan x.{{\log }_{e}}\sin x\] \[(0\cdot \infty \,\,form)\] \[\underset{x\to 0}{\mathop{\lim }}\,\frac{{{\log }_{e}}\sin x}{\cot x}\left( \frac{\infty }{\infty }form \right)\] \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{\cot x}{-\cos e{{c}^{2}}x}\] [Using L?Hospital?s rule] \[=\underset{x\to 0}{\mathop{\lim }}\,(-cosx.sinx)=0.\]
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