नीचे दो कथन दिए गए हैं:कथन I: lim _{x rightar | vedclass

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(D) For Statement $I$:Using Taylor series expansions:$	an ^{-1} x = x - \frac{x^3}{3} + \frac{x^5}{5} - \dots$$\log _e \sqrt{\frac{1+x}{1-x}} = \frac

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दोनों कथन $I$ और कथन $II$ सत्य हैं

Vedclass · Answer

(D) For Statement $I$:Using Taylor series expansions:$	an ^{-1} x = x - \frac{x^3}{3} + \frac{x^5}{5} - \dots$$\log _e \sqrt{\frac{1+x}{1-x}} = \frac{1}{2} [\ln(1+x) - \ln(1-x)] = \frac{1}{2} [(x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \frac{x^5}{5}) - (-x - \frac{x^2}{2} - \frac{x^3}{3} - \frac{x^4}{4} - \frac{x^5}{5})] = x + \frac{x^3}{3} + \frac{x^5}{5} + \dots$Substituting these into the limit:$\lim _{x ightarrow 0} \frac{(x - \frac{x^3}{3} + \frac{x^5}{5}) + (x + \frac{x^3}{3} + \frac{x^5}{5}) - 2x}{x^5} = \lim _{x ightarrow 0} \frac{2x + \frac{2x^5}{5} - 2x}{x^5} = \frac{2}{5}$.Thus,Statement $I$ is true.For Statement $II$:Let $L = \lim _{x ightarrow 1} x^{\frac{2}{1-x}}$. This is a $1^\infty$ form.$L = e^{\lim _{x ightarrow 1} (x-1) \cdot \frac{2}{1-x}} = e^{\lim _{x ightarrow 1} -2} = e^{-2} = \frac{1}{e^2}$.Thus,Statement $II$ is true.

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