Question

If \( f(x) = \left(\frac{1+x}{1-x}\right)^{\frac{1}{x}} \) is continuous at \( x = 0 \), then \( f(0) \) is:

• A. \( e^{\frac{1}{2}} \)
• B. \( e^2 \)
• C. \( e^{-2} \)
• D. \( e^{-\frac{1}{2}} \)

Hint:

For limits of the form \( \left(1 + x\right)^{\frac{1}{x}} \), use the approximation \( \ln(1 + x) \approx x \) for small \( x \).

Answer 1

The Correct Option is B

Step 1: Establish the Limit Condition For \( f(x) \) to be continuous at \( x = 0 \), we need: \[ \lim_{x \to 0} f(x) = f(0). \] Given the function: \[ f(x) = \left(\frac{1+x}{1-x}\right)^{\frac{1}{x}}, \] we evaluate: \[ \lim_{x \to 0} \left(\frac{1+x}{1-x}\right)^{\frac{1}{x}}. \] 
Step 2: Apply Natural Logarithm Define: \[ L = \lim_{x \to 0} \left(\frac{1+x}{1-x}\right)^{\frac{1}{x}}. \] Taking the logarithm: \[ \ln L = \lim_{x \to 0} \frac{1}{x} \ln \left(\frac{1+x}{1-x}\right). \] Using the approximations: \[ \ln(1 + x) \approx x, \quad \ln(1 - x) \approx -x \quad \text{for small } x, \] we approximate: \[ \ln \left(\frac{1+x}{1-x}\right) = \ln(1 + x) - \ln(1 - x) \approx x - (-x) = 2x. \] Thus, the equation simplifies to: \[ \ln L = \lim_{x \to 0} \frac{1}{x} \cdot 2x = 2. \] 
Step 3: Determine the Final Value Since \( L = e^{\ln L} \), we obtain: \[ L = e^2. \] Thus, the required value is: 

\[Option (2): e^2.\]