1.
If α=lim⁡x→0x⋅2x−x1−cos⁡x\alpha=\displaystyle\lim_{x\to0}\frac{x\cdot2^x-x}{1-\cos x} and β=lim⁡x→0x⋅2x−x1+x2−1−x2\beta=\displaystyle\lim_{x\to0}\frac{x\cdot2^x-x}{\sqrt{1+x^2}-\sqrt{1-x^2}}, then