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: $\lim_{x \to 1} \left( \frac{2}{x^{1-x}} \right) = \frac{1}{e^2}$ In the light of the above statements, choose the correct answer from the options given below

• A. Both Statement I and Statement II are false
• B. Statement I is false but Statement II is true
• C. Both Statement I and Statement II are true
• D. Statement I is true but Statement II is false

Hint:

- For limit evaluations near 0, Taylor series expansions are often useful - For \( 1^\infty \) forms, use logarithmic transformation - Remember L'Hôpital's rule for indeterminate forms - Verify both statements independently before choosing the option

Answer 1

The Correct Option is C

Verification of Statement I: Using Taylor series expansions about \( x = 0 \): \[ \tan^{-1}x = x - \frac{x^3}{3} + \frac{x^5}{5} - \cdots \] \[ \log_e \sqrt{\frac{1+x}{1-x}} = \frac{1}{2} \left( \log(1+x) - \log(1-x) \right) = x + \frac{x^3}{3} + \frac{x^5}{5} + \cdots \] Substituting into the limit: \[ \frac{(x - \frac{x^3}{3} + \frac{x^5}{5}) + (x + \frac{x^3}{3} + \frac{x^5}{5}) - 2x}{x^5} = \frac{\frac{2x^5}{5}}{x^5} = \frac{2}{5} \] Thus, Statement I is true.

Verification of Statement II: Let \( y = \frac{2}{x^{1-x}} \).
Taking natural log: \[ \ln y = \frac{2}{1-x} \ln x \] Using L'Hôpital's rule as \( x \to 1 \): \[ \lim_{x \to 1} \frac{2\ln x}{1-x} = \lim_{x \to 1} \frac{2/x}{-1} = -2 \] Thus: \[ \lim_{x \to 1} y = e^{-2} = \frac{1}{e^2} \] Therefore, Statement II is true.