If $\alpha =\underset{x\to 0}{\lim }\frac{x.2^x-x}{1-\cos x}$ and $\beta =\underset{x\to 0}{\lim }\frac{x.2^x-x}{\sqrt{1+x^2}-\sqrt{1-x^2}}$ then

$$\begin{gathered} \mathrm{\alpha }=\underset{\mathrm{x}\to 0}{\lim }\frac{\mathrm{x}.2^{\mathrm{x}}-\mathrm{x}}{1-\mathrm{cosx}} \\ =\underset{\mathrm{x}\to 0}{\lim } \frac{\mathrm{x}(2^{\mathrm{x}}-1)}{1-\cos \mathrm{x}} \\ =\underset{\mathrm{x}\to 0}{\lim } \frac{{\displaystyle \frac{2^{\mathrm{x}}-1}{\mathrm{x}}}}{{\displaystyle \frac{1-\cos \mathrm{x}}{{\mathrm{x}}^2}}} \\ =\frac{{\log }^2}{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}=2 \log 2 \end{gathered}$$

$\mathrm{\beta }=\underset{\mathrm{x}\to 0}{\lim }\frac{\mathrm{x}.2^{\mathrm{x}}-\mathrm{x}}{\sqrt{1+{\mathrm{x}}^2}-\sqrt{1-{\mathrm{x}}^2}}$
$\mathrm{\beta }=\underset{\mathrm{x}\to 0}{\lim } \frac{\mathrm{x}(2^{\mathrm{x}}-1)}{1+{\mathrm{x}}^2-1+{\mathrm{x}}^2} \left(\sqrt{1+{\mathrm{x}}^2}+\sqrt{1-{\mathrm{x}}^2}\right)$
$$\begin{gathered} =\underset{\mathrm{x}\to 0}{\lim }\frac{\mathrm{x}(2^{\mathrm{x}}-1)}{2{\mathrm{x}}^2} \left(\sqrt{1+{\mathrm{x}}^2}+\sqrt{1-{\mathrm{x}}^2}\right) \\ =\underset{\mathrm{x}\to 0}{\lim } \frac{2^{\mathrm{x}}-1}{2\mathrm{x}} \left(\sqrt{1+{\mathrm{x}}^2}+\sqrt{1-{\mathrm{x}}^2}\right) \\ =\frac{\log 2}{2}\left(2\right) \\ =\log 2 \\ \mathrm{\alpha }=2\mathrm{\beta } \end{gathered}$$