यदि alpha=lim _{x rightarrow 0} frac{x cdot 2 | vedclass

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(C) $\alpha$ ज्ञात करने के लिए: $\alpha = \lim _{x \rightarrow 0} \frac{x(2^x-1)}{1-\cos x} = \lim _{x \rightarrow 0} \frac{x(2^x-1)}{2 \sin^2(x/2)}$

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$\alpha=2 \beta$

Vedclass · Answer

(C) $\alpha$ ज्ञात करने के लिए: $\alpha = \lim _{x \rightarrow 0} \frac{x(2^x-1)}{1-\cos x} = \lim _{x \rightarrow 0} \frac{x(2^x-1)}{2 \sin^2(x/2)}$ $= \lim _{x \rightarrow 0} \frac{1}{2} \cdot \frac{2^x-1}{x} \cdot \frac{x^2}{\sin^2(x/2)} \cdot \frac{1}{x} = \lim _{x \rightarrow 0} \frac{1}{2} \cdot \frac{2^x-1}{x} \cdot \frac{1}{(\frac{\sin(x/2)}{x/2})^2 \cdot \frac{1}{4}} = \frac{1}{2} \cdot \ln 2 \cdot 4 = 2 \ln 2$ अतः,$\alpha = 2 \ln 2$ ... $(i)$ $\beta$ ज्ञात करने के लिए: $\beta = \lim _{x \rightarrow 0} \frac{x(2^x-1)}{\sqrt{1+x^2}-\sqrt{1-x^2}}$ हर का परिमेयकरण करने पर: $\beta = \lim _{x \rightarrow 0} \frac{x(2^x-1)(\sqrt{1+x^2}+\sqrt{1-x^2})}{(1+x^2)-(1-x^2)} = \lim _{x \rightarrow 0} \frac{x(2^x-1)(\sqrt{1+x^2}+\sqrt{1-x^2})}{2x^2}$ $= \lim _{x \rightarrow 0} \frac{1}{2} \cdot \frac{2^x-1}{x} \cdot (\sqrt{1+x^2}+\sqrt{1-x^2}) = \frac{1}{2} \cdot \ln 2 \cdot (1+1) = \ln 2$ अतः,$\beta = \ln 2$ ... $(ii)$ $(i)$ और $(ii)$ से,हमें $\alpha = 2\beta$ प्राप्त होता है।

यदि $\alpha=\lim _{x \rightarrow 0} \frac{x \cdot 2^x-x}{1-\cos x}$ और $\beta=\lim _{x \rightarrow 0} \frac{x \cdot 2^x-x}{\sqrt{1+x^2}-\sqrt{1-x^2}}$ है,तो

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