Given below are two statements :
Statement I : \(\lim _{x\to 0}\left(\frac{{\tan }^{-1}x+{\log }_{e}\sqrt{\frac{1+x}{1-x}}-2x}{{x}^{5}}\right)=\frac{2}{5}\)
Statement II : \(\lim _{x\to 1}\left({x}^{\frac{2}{1-x}}\right)=\frac{1}{{e}^{2}}\)
In the light of the above statements, choose the correct answer from the options given below :

[JEE Main 2025, 8 Apr (Shift 1)]

Question

Given below are two statements :
Statement I : \(\lim _{x\to 0}\left(\frac{{\tan }^{-1}x+{\log }_{e}\sqrt{\frac{1+x}{1-x}}-2x}{{x}^{5}}\right)=\frac{2}{5}\)
Statement II : \(\lim _{x\to 1}\left({x}^{\frac{2}{1-x}}\right)=\frac{1}{{e}^{2}}\)
In the light of the above statements, choose the correct answer from the options given below :

[JEE Main 2025, 8 Apr (Shift 1)]

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Given below are two statements : Statement I : \(\lim _{x\to 0}\left(\frac{{\tan }^{-1}x+{\log }_{e}\sqrt{\frac{1+x}{1-x…

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