Question

$\lim_{x \to 0} \left( \frac{1}{x} \ln \left( \frac{\sqrt{1 + x}}{\sqrt{1 - x}} \right) \right)$ is

• A. 1/2
• B. 0
• C. 1
• D. does not exist

Answer 1

The Correct Option is C

We are tasked with evaluating the limit: \[ \lim_{x \to 0} \left( \frac{1}{x} \ln \left( \frac{\sqrt{1 + x}}{\sqrt{1 - x}} \right) \right) \] Step 1: Simplify the expression inside the logarithm First, we simplify the argument of the logarithmic function: \[ \frac{\sqrt{1 + x}}{\sqrt{1 - x}} = \sqrt{\frac{1 + x}{1 - x}} \] Thus, the given limit becomes: \[ \lim_{x \to 0} \frac{1}{x} \ln \left( \sqrt{\frac{1 + x}{1 - x}} \right) \] Since \( \ln \left( \sqrt{A} \right) = \frac{1}{2} \ln A \), we have: \[ \lim_{x \to 0} \frac{1}{x} \cdot \frac{1}{2} \ln \left( \frac{1 + x}{1 - x} \right) \] This simplifies to: \[ \frac{1}{2} \lim_{x \to 0} \frac{1}{x} \ln \left( \frac{1 + x}{1 - x} \right) \] Step 2: Expand the logarithmic expression Now, let's expand the logarithmic function for small \( x \). Using the first-order Taylor expansion for \( \ln(1 + x) \), we get: \[ \ln(1 + x) \approx x \quad \text{and} \quad \ln(1 - x) \approx -x \] Thus, \[ \ln \left( \frac{1 + x}{1 - x} \right) = \ln(1 + x) - \ln(1 - x) \approx x - (-x) = 2x \] Step 3: Substitute into the limit expression Now substitute the approximation \( \ln \left( \frac{1 + x}{1 - x} \right) \approx 2x \) into the limit expression: \[ \frac{1}{2} \lim_{x \to 0} \frac{1}{x} \cdot 2x \] This simplifies to: \[ \frac{1}{2} \lim_{x \to 0} 2 = 1 \] Conclusion The value of the limit is: \[ \boxed{1} \]