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

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

Vedclass · Accepted Answer

$\alpha = 2\beta$

Vedclass · Answer

(B) $\alpha = \lim_{x ightarrow 0} \frac{x(2^x - 1)}{1 - \cos x}$ માટે. $L'	ext{Hôpital's rule}$ લાગુ પાડતા:$\alpha = \lim_{x ightarrow 0} \frac{(2^x - 1) + x \cdot 2^x \ln 2}{\sin x}$.ફરીથી $L'	ext{Hôpital's rule}$ લાગુ પાડતા:$\alpha = \lim_{x ightarrow 0} \frac{2^x \ln 2 + 2^x \ln 2 + x \cdot 2^x (\ln 2)^2}{\cos x} = \frac{\ln 2 + \ln 2 + 0}{1} = 2 \ln 2$.$\beta = \lim_{x ightarrow 0} \frac{x(2^x - 1)}{\sqrt{1+x^2} - \sqrt{1-x^2}}$ માટે. છેદનું સંમેયીકરણ કરતા:$\beta = \lim_{x ightarrow 0} \frac{x(2^x - 1)(\sqrt{1+x^2} + \sqrt{1-x^2})}{(1+x^2) - (1-x^2)} = \lim_{x ightarrow 0} \frac{x(2^x - 1)(\sqrt{1+x^2} + \sqrt{1-x^2})}{2x^2} = \lim_{x ightarrow 0} \frac{(2^x - 1)}{x} \cdot \frac{(\sqrt{1+x^2} + \sqrt{1-x^2})}{2}$.કારણ કે $\lim_{x ightarrow 0} \frac{2^x - 1}{x} = \ln 2$,તેથી $\beta = \ln 2 \cdot \frac{1+1}{2} = \ln 2$.આમ,$\alpha = 2\beta$.

જો $\alpha = \lim_{x \rightarrow 0} \frac{x \cdot 2^x - x}{1 - \cos x}$ અને $\beta = \lim_{x \rightarrow 0} \frac{x \cdot 2^x - x}{\sqrt{1+x^2} - \sqrt{1-x^2}}$,હોય તો

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