Question

$\underset{x\to 0}{\lim }\frac{x\left(2^x-1\right)}{1-\cos x}$

Answer 1

$\underset{x\to 0}{\lim }\left[\frac{x\left(2^x-1\right)}{1-\cos x}\right]$

Dividing the numerator and the denominator by x²:

$$\begin{gathered} =\underset{x\to 0}{\lim }\left[\left(\frac{2^x-1}{x}\right)\times \frac{x^2}{1-\cos x}\right] \\ =\underset{x\to 0}{\lim }\left[\left(\frac{2^x-1}{x}\right)\times \frac{x^2}{2{\sin }^2{\displaystyle \frac{x}{2}}}\right] \\ =\underset{x\to 0}{\lim }\left[\left(\frac{2^x-1}{x}\right)\times \frac{{\displaystyle \frac{x^2}{4}}\times 4}{2{\sin }^2\left({\displaystyle \frac{x}{2}}\right)}\right] \\ =\left(\log 2\right)\times \frac{4}{2}\times \frac{1}{1^2} \\ =2\log 2 \\ =\log {\left(2\right)}^2 \\ =\log 4 \end{gathered}$$