Question

Consider two statements: 
Statement 1: $ \lim_{x \to 0} \frac{\tan^{-1} x + \ln \left( \frac{1+x}{1-x} \right) - 2x}{x^5} = \frac{2}{5} $ 
Statement 2: $ \lim_{x \to 1} x \left( \frac{2}{1-x} \right) = e^2 \; \text{and can be solved by the method}  \lim_{x \to 1} \frac{f(x)}{g(x) - 1} $

• A. Only Statement 1 is true
• B. Only Statement 2 is true
• C. Both Statement 1 and Statement 2 are true
• D. Both Statement 1 and Statement 2 are false

Hint:

For solving limits, use series expansions and L'Hopital's Rule to simplify complicated expressions, making evaluation easier.

Answer 1

The Correct Option is A

For Statement 1, we evaluate the given limit using series expansions: \[ \lim_{x \to 0} \frac{\tan^{-1} x + \ln \left( \frac{1+x}{1-x} \right) - 2x}{x^5} \] Expanding the terms: \[ \tan^{-1} x = x - \frac{x^3}{3} + \frac{x^5}{5} + \cdots \] \[ \ln \left( \frac{1+x}{1-x} \right) = 2x + \frac{x^3}{3} + \frac{2x^5}{5} + \cdots \] Substituting and simplifying: \[ \frac{x - \frac{x^3}{3} + \frac{x^5}{5} + 2x + \frac{x^3}{3} + \frac{2x^5}{5} - 2x}{x^5} \] \[ = \frac{\frac{2x^5}{5}}{x^5} = \frac{2}{5} \] Thus, Statement 1 is true. For Statement 2, we evaluate the given limit: \[ \lim_{x \to 1} x \left( \frac{2}{1-x} \right) \] This does not give \( e^2 \) as claimed.
The expression diverges and does not yield the desired result. Thus, Statement 2 is false.