A) \[\frac{1}{e}\]
B) 1
C) e
D) \[{{e}^{2}}\]
E) \[\frac{1}{{{e}^{2}}}\]
Correct Answer: B
Solution :
\[\underset{x\to 0}{\mathop{\lim }}\,{{\left\{ \frac{1+\tan x}{1+\sin x} \right\}}^{\cos ecx}}\] \[=\underset{x\to 0}{\mathop{\lim }}\,\left\{ \frac{{{\left[ {{\left( 1+\frac{\sin x}{\cos x} \right)}^{\frac{\cos x}{\sin x}}} \right]}^{1/\cos x}}}{{{(1+\sin x)}^{1/\sin x}}} \right\}\] \[=\frac{{{e}^{\underset{x\to 0}{\mathop{\lim }}\,1/\cos x}}}{e}=\frac{e}{e}=1\]
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