We have,
$$
\begin{aligned}
& \alpha=\lim _{x \rightarrow 0} \frac{x \cdot 2^x-x}{1-\cos x} \\
& {\left[\frac{0}{0} \text { form }\right]} \\
& =\lim _{x \rightarrow 0} \frac{1 \cdot 2^x+x \cdot 2^x \log 2-1}{\sin x} \\
& {\left[\frac{0}{0} \text { form }\right]} \\
& =\lim _{x \rightarrow 0} \frac{2^x \log 2+2^x \log 2+x\left(\log 2^2 2^x\right.}{\cos x} \\
& =\lim _{x \rightarrow 0} \frac{2 \cdot 2^x \log 2+2^x \cdot x\left(\log 2^2\right.}{\cos x} \\
& \Rightarrow \alpha=2 \cdot \log 2 \\
& \text { Again, } \beta=\lim _{x \rightarrow 0} \frac{x \cdot 2^x-x}{\sqrt{1+x^2}-\sqrt{1-x^2}} \\
& {\left[\frac{0}{0} \text { form }\right]} \\
& =\lim _{x \rightarrow 0} \frac{x 2^x-x}{\sqrt{1+x^2}-\sqrt{1-x^2}} \times \frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}+\sqrt{1-x^2}} \\
& {\left[\frac{0}{0} \text { form }\right]} \\
&
\end{aligned}
$$


$$
\begin{aligned}
& =\lim _{x \rightarrow 0} \frac{\left(x 2^x-x\right)\left(\sqrt{1+x^2}+\sqrt{1-x^2}\right)}{1+x^2-1+x^2} \quad\left[\frac{0}{0} \text { form }\right] \\
& =\lim _{x \rightarrow 0} \frac{\left(z^x-1\right)\left[\sqrt{1+x^2}+\sqrt{1-x^2}\right]}{2 x} \quad\left[\frac{0}{0} \text { form }\right] \\
& =\lim _{x \rightarrow 0} \frac{2^x \log 2\left(\sqrt{1+x^2}+\sqrt{1-x^2}\right)+\left(z^x-1\right)}{2} \quad\left[\frac{2 x}{\sqrt{1+x^2}}-\frac{2 x}{\sqrt{1-x^2}}\right] \\
& \Rightarrow \beta=\frac{\log 2(1+1)+0}{2}=\log 2 \\
& \therefore \quad \alpha=2 \beta \quad
\end{aligned}
$$