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

KCETKCET 2006Limits and Derivatives

Solution:

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