A) \[{{\log }_{e}}2\]
B) \[{{\log }_{e}}\sqrt{2}\]
C) \[{{\log }_{e}}4\]
D) 2
E) \[\frac{1}{2}\]
Correct Answer: C
Solution :
\[\underset{x\to 0}{\mathop{\lim }}\,\left[ \frac{{{2}^{x}}-1}{\sqrt{1+x}-1} \right]\] \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{\frac{{{2}^{x}}\log 2}{1}}{2\sqrt{1+x}}\] (by LHospitals rule) \[=2\text{ }log\text{ }2=\text{ }log\text{ }4.\]
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