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: 

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

Answer 2

To determine \( f(0) \) when \( f(x) = \left(\frac{1+x}{1-x}\right)^{\frac{1}{x}} \) is continuous at \( x = 0 \), we need to find the limit of \( f(x) \) as \( x \) approaches 0. We apply the expansion of the natural logarithm around 0.
First, take the natural logarithm:
\( \ln(f(x)) = \frac{1}{x} \ln\left(\frac{1+x}{1-x}\right) \)
Using the logarithm property \( \ln(a/b) = \ln(a) - \ln(b) \), we get:
\( \ln(f(x)) = \frac{1}{x}(\ln(1+x) - \ln(1-x)) \)
Apply the series expansions: \( \ln(1+x) \approx x - \frac{x^2}{2} + \frac{x^3}{3} + \cdots \) and \( \ln(1-x) \approx -x - \frac{x^2}{2} - \frac{x^3}{3} + \cdots \). So,
\( \ln(1+x) - \ln(1-x) \approx (x - \frac{x^2}{2} + \frac{x^3}{3}) - (-x - \frac{x^2}{2} - \frac{x^3}{3}) \).
Combine terms:
\( 2x + \frac{2x^3}{3} \)
Divide by \( x \):
\( \frac{1}{x}(2x + \frac{2x^3}{3}) \rightarrow 2 + \frac{2x^2}{3} \)
As \( x \to 0 \), the expression tends to 2. Hence, the limit of \( \ln(f(x)) \) as \( x \to 0 \) is 2, so the limit of \( f(x) \) as \( x \to 0 \) is \( e^2 \). Thus, if \( f(x) \) is continuous at \( x = 0 \), then \( f(0) = e^2 \).