Question

Use the formula $ lim_{ x \to 0 } \frac{a^x - 1}{ x} = log_e \, a, $ to find
$ lim_{ x \to 0 } \frac{ 2^x - 1}{ (1 + x)^{1/2} - 1 } $.

Answer 1

$ lim_{ x \to 0 } \frac{ 2^x - 1}{ \sqrt {1 + x} - 1} \times \frac{ \sqrt {1 + x} + 1}{ \sqrt{1 + x} + 1} = lim_{ x \to 0 } \frac{ (2^x - 1) \, ( \sqrt{1 + x} + 1)}{ x} $
= $log_e $ (2) . (2)
= 2 $ log_e 2 = log_e 4 $.