A) 0
B) 1
C) 1/e
D) none of these
Correct Answer: C
Solution :
Let \[y=\underset{x\to 0}{\mathop{\lim }}\,{{(\cos ecx)}^{1/\log x}}\] On taking log on both sides, we get \[\log y=\underset{x\to 0}{\mathop{\lim }}\,\frac{\log \cos ecx}{\log x}\left( \frac{\infty }{\infty }form \right)\] \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{-\cot x}{1/x}\] (by LHospitals rule) \[=-\underset{x\to 0}{\mathop{\lim }}\,\frac{x}{\tan x}=-1\] \[\therefore \] \[\log y=-1\] \[\Rightarrow \] \[y={{e}^{-1}}=\frac{1}{e}\]
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