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