A) log 5
B) 0
C) 1
D) \[2\text{ }log\text{ }5\]
Correct Answer: A
Solution :
\[\underset{x\to 0}{\mathop{\lim }}\,\frac{{{5}^{x}}-{{5}^{-x}}}{2x}\] Applying L- Hospitals rule \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{{{5}^{x}}\log 5+{{5}^{-x}}\log 5}{2}\] \[=\frac{\log 5+\log 5}{2}=\log 5\]You need to login to perform this action.
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