A) e
B) \[{{e}^{2}}\]
C) \[\frac{1}{2}\]
D) 2
Correct Answer: D
Solution :
\[\underset{x\to 0}{\mathop{\lim }}\,\frac{2}{x}\log (1+x)=\underset{x\to 0}{\mathop{\lim }}\,2\log {{(1+x)}^{\frac{1}{x}}}\] \[=\underset{x\to 0}{\mathop{\lim }}\,2{{\log }_{e}}e=2\] \[\left\{ \because \underset{x\to 0}{\mathop{\lim }}\,{{(1+x)}^{\frac{1}{x}}}={{\log }_{e}}e=1 \right\}\] Trick : Using L Hospital?s rule.
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