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 :

Let the given limit in Statement I be $$L_1$$:

$$L_1 = \lim_{x \to 0} \left(\frac{\tan^{-1}x + \log_{e}\sqrt{\frac{1+x}{1-x}} - 2x}{x^5}\right)$$

Using Maclaurin series expansions up to the $$x^5$$ term (terms with powers greater than 5 are ignored as division by $$x^5$$ gives them positive powers which vanish as $$x \to 0$$):

$$\tan^{-1}x = x - \frac{x^3}{3} + \frac{x^5}{5}$$

$$\log_e(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \frac{x^5}{5}$$

$$\log_e(1-x) = -x - \frac{x^2}{2} - \frac{x^3}{3} - \frac{x^4}{4} - \frac{x^5}{5}$$

Subtracting these expressions, we get:

$$\log_e(1+x) - \log_e(1-x) = \left(x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \frac{x^5}{5}\right) - \left(-x - \frac{x^2}{2} - \frac{x^3}{3} - \frac{x^4}{4} - \frac{x^5}{5}\right)$$

$$\log_e(1+x) - \log_e(1-x) = 2x + \frac{2x^3}{3} + \frac{2x^5}{5}$$

Thus, the second term becomes:

$$\log_{e}\sqrt{\frac{1+x}{1-x}} = \frac{1}{2}\left[\log_e(1+x) - \log_e(1-x)\right] = x + \frac{x^3}{3} + \frac{x^5}{5}$$

Substituting these into the numerator:

$$\left(x - \frac{x^3}{3} + \frac{x^5}{5}\right) + \left(x + \frac{x^3}{3} + \frac{x^5}{5}\right) - 2x = \frac{2x^5}{5}$$

Evaluating the limit:

$$L_1 = \lim_{x \to 0} \frac{\frac{2}{5}x^5}{x^5} = \frac{2}{5}$$

Thus, Statement I is true.

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Let the given limit in Statement II be $$L_2$$:

$$L_2 = \lim_{x \to 1} \left(x^{\frac{2}{1-x}}\right)$$

Taking natural logarithm on both sides:

$$\log_e L_2 = \lim_{x \to 1} \frac{2 \log_e x}{1-x}$$

As the limi is in the form $$\frac{0}{0}$$, we apply L'Hopital's Rule that is we differentiate the numerator and denominator to get:

$$\log_e L_2 = \lim_{x \to 1} \frac{\frac{2}{x}}{-1} = -2$$

$$L_2 = e^{-2} = \frac{1}{e^2}$$

Thus, Statement II is true.

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Therefore, both Statement I and Statement II are true.