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

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(B) For $\alpha = \lim_{x ightarrow 0} \frac{x(2^x - 1)}{1 - \cos x}$. Applying $L'	ext{Hôpital's rule}$:$\alpha = \lim_{x ightarrow 0} \frac{(2^

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$\alpha = 2\beta$

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(B) For $\alpha = \lim_{x ightarrow 0} \frac{x(2^x - 1)}{1 - \cos x}$. Applying $L'	ext{Hôpital's rule}$:$\alpha = \lim_{x ightarrow 0} \frac{(2^x - 1) + x \cdot 2^x \ln 2}{\sin x}$.Applying $L'	ext{Hôpital's rule}$ again:$\alpha = \lim_{x ightarrow 0} \frac{2^x \ln 2 + 2^x \ln 2 + x \cdot 2^x (\ln 2)^2}{\cos x} = \frac{\ln 2 + \ln 2 + 0}{1} = 2 \ln 2$.For $\beta = \lim_{x ightarrow 0} \frac{x(2^x - 1)}{\sqrt{1+x^2} - \sqrt{1-x^2}}$. Rationalizing the denominator:$\beta = \lim_{x ightarrow 0} \frac{x(2^x - 1)(\sqrt{1+x^2} + \sqrt{1-x^2})}{(1+x^2) - (1-x^2)} = \lim_{x ightarrow 0} \frac{x(2^x - 1)(\sqrt{1+x^2} + \sqrt{1-x^2})}{2x^2} = \lim_{x ightarrow 0} \frac{(2^x - 1)}{x} \cdot \frac{(\sqrt{1+x^2} + \sqrt{1-x^2})}{2}$.Since $\lim_{x ightarrow 0} \frac{2^x - 1}{x} = \ln 2$,we have $\beta = \ln 2 \cdot \frac{1+1}{2} = \ln 2$.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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