Limits Continuity And Differentiability

Use the formula \(\mathop {\lim }\limits_{x \to 0} {{{a^x} - 1} \over x} = {\log _e}a\), to compute \(\mathop {\lim }\limits_{x \to 0} {{{2^x} - 1} \over {\sqrt {1 + x} - 1}}\).