A) \[\log 5\]
B) 0
C) 1
D) \[2\,\log 5\]
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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