If alpha = lim_{x rightarrow 0} frac{x cdot 2 | vedclass

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(B) Given $\alpha = \lim_{x \rightarrow 0} \frac{x(2^x - 1)}{1 - \cos x}$.Applying $L'H\hat{o}pital's$ rule:$\alpha = \lim_{x \rightarrow 0} \frac{2^x

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$\alpha = 2 \beta$

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(B) Given $\alpha = \lim_{x \rightarrow 0} \frac{x(2^x - 1)}{1 - \cos x}$.Applying $L'H\hat{o}pital's$ rule:$\alpha = \lim_{x \rightarrow 0} \frac{2^x - 1 + x \cdot 2^x \ln 2}{\sin x}$.Applying $L'H\hat{o}pital's$ rule again:$\alpha = \lim_{x \rightarrow 0} \frac{2^x \ln 2 + 2^x \ln 2 + x \cdot 2^x (\ln 2)^2}{\cos x} = \frac{\ln 2 + \ln 2 + 0}{1} = 2 \ln 2$.Now,$\beta = \lim_{x \rightarrow 0} \frac{x(2^x - 1)}{\sqrt{1 + x^2} - \sqrt{1 - x^2}}$.Rationalizing the denominator:$\beta = \lim_{x \rightarrow 0} \frac{x(2^x - 1)(\sqrt{1 + x^2} + \sqrt{1 - x^2})}{(1 + x^2) - (1 - x^2)} = \lim_{x \rightarrow 0} \frac{x(2^x - 1)(\sqrt{1 + x^2} + \sqrt{1 - x^2})}{2x^2}$.$\beta = \frac{1}{2} \lim_{x \rightarrow 0} \frac{2^x - 1}{x} \cdot (\sqrt{1 + x^2} + \sqrt{1 - x^2}) = \frac{1}{2} \cdot \ln 2 \cdot (1 + 1) = \ln 2$.Thus,$\alpha = 2 \beta$.

If $\alpha = \lim_{x \rightarrow 0} \frac{x \cdot 2^x - x}{1 - \cos x}$ and $\beta = \lim_{x \rightarrow 0} \frac{x \cdot 2^x - x}{\sqrt{1 + x^2} - \sqrt{1 - x^2}}$,then

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