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