If α = lim x → 0 x · 2 x - x 1 - cos x and β = lim x → 0 x · 2 x - x 1 + x 2…

It is given that, $\alpha =\underset{x\to 0}{\lim }\frac{x·2^x-x}{1-\cos x}$

Apply L hospital rule, $a=\underset{x\to 0}{\lim }\frac{2^x+x·2^x\log 2-1}{\sin x}$

Again, apply L hospital rule,

$\alpha =\underset{x\to 0}{\lim }\frac{2^x\log 2+2^x\log 2+x·2^x(\log 2{)}^2}{\cos x}$

$\alpha =\log 2+\log 2$

$\alpha =2\log 2 \dots \left(\mathrm{i}\right)$

Now

$\beta =\underset{x\to 0}{\lim }\frac{x·2^x-x}{\sqrt{1+x^2}-\sqrt{1-x^2}}$

Apply L hospital rule,

$\beta =\underset{x\to 0}{\lim }\frac{2^x+x·2^x\log 2-1}{{\displaystyle \frac{x}{\sqrt{1+x^2}}}+\frac{x}{\sqrt{1-x^2}}}$

Again, apply L hospital rule,

$\beta =\underset{x\to 0}{\lim }\frac{2^x\log 2+2^x\log 2+x·2^x(\log 2{)}^2}{{\displaystyle \frac{\sqrt{1+x^2}-{\displaystyle \frac{x^2}{\sqrt{1+x^2}}}}{1+x^2}}+\frac{\sqrt{1-x^2}+\frac{x^2}{\sqrt{1-x^2}}}{1-x^2}}$

$\beta =\frac{2\log 2}{2}=\log \left(2\right) \dots \left(\mathrm{ii}\right)$

From equation $\left(\mathrm{i}\right)$ and $\left(\mathrm{ii}\right),$

$\alpha =2\beta$