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

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