A) \[\frac{1}{e}\]
B) \[1\]
C) \[e\]
D) \[{{e}^{2}}\]
Correct Answer: B
Solution :
Consider \[\underset{x\to 0}{\mathop{\lim }}\,\,{{\left\{ \frac{1+\tan x}{1+\sin x} \right\}}^{\cos ecx}}\] \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{{{\left[ {{\left( 1+\frac{\sin x}{\cos x} \right)}^{\frac{\cos x}{\sin x}}} \right]}^{1/\cos x}}}{{{(1+\sin x)}^{1/\sin x}}}\] \[[\because \,\,\tan x=\frac{\sin x}{\cos x}\,\,and\,\,\text{cosec}\,x=\frac{1}{\sin x}]\] We know, \[\underset{n\to 0}{\mathop{\lim }}\,{{\left( 1+\frac{1}{n} \right)}^{n}}=e\] \[\therefore \,\,\underset{x\to 0}{\mathop{\lim }}\,\,\,\frac{{{\left[ {{\left( 1+\frac{\sin x}{\cos x} \right)}^{\frac{\cos x}{\sin x}}} \right]}^{1/\cos x}}}{{{(1+\sin x)}^{1/\sin x}}}\] \[=\underset{x\to 0}{\mathop{\lim }}\,\,\,\frac{{{\left[ \left( 1+\frac{1}{\frac{\cos x}{\sin x}} \right) \right]}^{\frac{1}{\cos x}}}}{\left[ {{\left( 1+\frac{1}{\text{cosec x}} \right)}^{\text{cosec x}}} \right]}\] \[=\frac{{{e}^{\underset{x\to 0}{\mathop{\lim }}\,\,\,\frac{1}{\cos x}}}}{e}=\frac{e}{e}=1.\]
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