$\begin{aligned} \alpha & =\lim _{x \rightarrow 0} \frac{x \cdot 2^x-x}{1-\cos x} \\ \alpha & =\lim _{x \rightarrow 0} \frac{x\left(2^x-1\right)}{1-\cos x}=\lim _{x \rightarrow 0} \frac{x\left(2^x-1\right)}{2 \sin ^2 \frac{x}{2}} \\ & =\lim _{x \rightarrow 0} \frac{1}{2} \cdot \frac{2^x-1}{\sin ^2 \frac{x}{2}} \\ & =\frac{1}{2} \lim _{x \rightarrow 0} \frac{\frac{2^x-1}{x}}{\frac{\sin ^2 \frac{x}{2}}{x^2}} \quad[\because \text { divided by } x] \\ & =\frac{1}{2} \lim _{x \rightarrow 0} \frac{2^x-1}{\left(\frac{\sin ^2 \frac{1}{2}}{x}\right)^2}\end{aligned}$
$\begin{aligned} & =\frac{1}{2}\left[\lim _{x \rightarrow 0} \frac{2^x-1}{x} \times \frac{1}{\lim _{x \rightarrow 0}\left(\frac{\sin \frac{1}{2} x}{\frac{x}{2}}\right)^2 \times \frac{1}{4}}\right] \\ & {\left[\because \lim _{x \rightarrow 0} \frac{\sin x}{x}=1\right]} \\ & {\left[\because \lim _{x \rightarrow 0} \frac{a^x-1}{x}=\log a\right]} \\ & =\frac{1}{2} \times \log 2 \times \frac{1}{1 \times \frac{1}{4}}=\frac{1}{2} \log 2 \times 4 \\ & \end{aligned}$


$$
\begin{aligned}
& \text { Given, } \beta=\lim _{x \rightarrow 0} \frac{x \cdot 2^x-x}{\sqrt{1+x^2}-\sqrt{1-x^2}} \\
& \beta=\lim _{x \rightarrow 0} \frac{x\left(2^x-1\right)}{\sqrt{1+x^2}-\sqrt{1-x^2}}
\end{aligned}
$$

Rationalise the denominator
$$
\begin{aligned}
& =\lim _{x \rightarrow 0} \frac{x\left(2^x-1\right)}{\sqrt{1+x^2}-\sqrt{1-x^2}} \times \frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}+\sqrt{1-x^2}} \\
& =\lim _{x \rightarrow 0} \frac{x\left(2^x-1\right)\left(\sqrt{1+x^2}+\sqrt{\left.1-x^2\right)}\right.}{\left(1+x^2\right)-\left(1-x^2\right)} \\
& =\lim _{x \rightarrow 0} \frac{x\left(2^x-1\right)\left(\sqrt{1+x^2}+\sqrt{\left.1-x^2\right)}\right.}{2 x^2} \\
& =\lim _{x \rightarrow 0} \frac{1}{2} \times \frac{2^x-1}{x} \times\left(\sqrt{1+x^2}+\sqrt{1-x^2}\right) \\
& =\frac{1}{2} \times \log 2 \times 2 \\
& \beta=\log 2
\end{aligned}
$$

From Eqs. (i) and (ii), we get $\alpha=2 \beta$

Hence, option (c) is correct.