Question

Evaluate :- $\lim_{x \to 0} \dfrac{2^x-1}{\sqrt{1+x}-1}$

Answer 1

$\bf \to \: \green{{ \underline{answer}}} \\ \\ \sf \to \: \lim_{x \: \to \: 0} = \frac{2 {}^{x} - 1 }{ \sqrt{1 + x} - 1 } = log(4)$

$\bf \to \: \orange{{ \underline{step - \: by - \: step - \: explanation}}}$

$\sf \to \: \lim_{x \: \to \: 0} \: = \dfrac{2 {}^{x} - 1 }{ \sqrt{1 + x} - 1} \\ \\ \sf \to \: by \: using \: l \: hospital \: rule \\ \\ \sf \to \: \lim_{x \: \to \: 0} = \dfrac{2 {}^{x} log(2) }{(1 + x) {}^{ \dfrac{1}{2} } - 1 } \\ \\ \sf \to \: \lim_{x \: \to \: 0} = \dfrac{2 {}^{x} log(2) }{ \dfrac{1}{2}(1 + x) {}^{ \dfrac{ - 1}{2} } } \\ \\ \sf \to \: put \: x \: = 0 \: in \: equation \\ \\ \sf \to \: \lim_{x \: \to \: 0} = \frac{2 {}^{0} log(2) }{ \dfrac{1}{2} (1 + 0) {}^{ \dfrac{ - 1}{2} } } \\ \\ \sf \to \: \lim_{x \: \to \: 0} = 2 log(2) \\ \\ \sf \to \: \lim_{x \: \to \: 0} = log(4) = answer$

Answer 2

$\huge{\underline{\mathtt{\red{A}\pink{N}\green{S}\blue{W}\purple{E}\orange{R}}}}$

The final answer is log(4).