A) \[-1\]
B) \[0\]
C) \[1\]
D) \[2\]
Correct Answer: D
Solution :
[d] \[{{L}_{1}}=\underset{x\to 0}{\mathop{\lim }}\,\,\,{{(\sec x)}^{\cos ec\,x}}\] (\[{{1}^{\infty }}\]form) \[={{e}^{\underset{x\to 0}{\mathop{\lim }}\,\,\,\cos ecx\,(\sec \,x-1)}}={{e}^{\underset{x\to 0}{\mathop{\lim }}\,\,\,\frac{1-\cos x}{\sin 2x}}}={{e}^{0}}=1\] \[{{L}_{2}}=\underset{x\to 0}{\mathop{\lim }}\,\,\,{{(cot\,x)}^{\sin x}}\] (\[{{\infty }^{0}}\] form) \[\therefore \,\,\,\,\,\,\,\,{{\log }_{e}}{{L}_{2}}=\underset{x\to 0}{\mathop{\lim }}\,\,\,(\sin \,\,x).{{\log }_{e}}(\cot x)\] \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{{{\log }_{e}}(\cot x)}{\cos ec\,x}\] \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{\cos e{{c}^{2}}x}{\cot x}\frac{1}{\cos ec\,x\,\,\cot x}\] \[=\underset{x\to 0}{\mathop{\lim }}\,\,\,\frac{\sin x}{{{\cos }^{2}}x}=0\] \[\therefore \,\,\,\,\,\,\,\,\,\,{{L}_{2}}={{e}^{0}}=1\] So, \[{{L}_{2}}+{{L}_{2}}=1+1=2\]
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