A) 0
B) log 2
C) 4
D) \[{{e}^{4}}\]
E) log 4
Correct Answer: E
Solution :
\[\underset{x\to 0}{\mathop{\lim }}\,\frac{{{2}^{x}}-{{2}^{-x}}}{x}=\underset{x\to 0}{\mathop{\lim }}\,{{2}^{x}}\log 2+{{2}^{-x}}\log 2\] (by L Hospitals rule) \[=log\text{ }2+log\text{ }2=log\text{ }4\] Since, Function is continuous at\[x=0\]. \[\therefore \] \[f(0)=\underset{x\to 0}{\mathop{\lim }}\,\frac{{{2}^{x}}-{{2}^{-x}}}{x}\] \[\Rightarrow \] \[f(0)=\log 4\]You need to login to perform this action.
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