Question

The value of $\displaystyle\lim_{x \to 0} \frac {5^x-5^{-x}}{2x}=$ is :

• A. 2 log 5
• B. 1
• C. 0
• D. log 5

Answer 1

Answer: (D)

log 5

Answer 2

$\displaystyle \lim _{x \rightarrow 0} \frac{5^{x}-5^{-x}}{2 x}$
Applying L- Hospital's rule
$=\displaystyle\lim _{x \rightarrow 0} \frac{5^{x} \log 5+5^{-x} \log 5}{2}$
$=\frac{\log 5+\log 5}{2}=\log 5$