Question

If the function \[ f(x) = \frac{\sqrt{1+x} - 1}{x} \] is continuous at \( x = 0 \), then \( f(0) \) is:

• A. \( -\frac{1}{2} \)
• B. \( \frac{1}{3} \)
• C. \( \frac{1}{2} \)
• D. \( -\frac{1}{3} \)

Hint:

For limits involving square roots, multiply by the conjugate to simplify.

Answer 1

The Correct Option is C

Evaluating the limit To check continuity at \( x = 0 \): \[ \lim_{x \to 0} \frac{\sqrt{1+x} - 1}{x}. \] Multiplying numerator and denominator by the conjugate: \[ \lim_{x \to 0} \frac{(\sqrt{1+x} - 1)(\sqrt{1+x} + 1)}{x(\sqrt{1+x} + 1)}. \] Since \( (\sqrt{1+x} - 1)(\sqrt{1+x} + 1) = 1+x -1 = x \), we simplify: \[ \lim_{x \to 0} \frac{x}{x(\sqrt{1+x} + 1)} = \lim_{x \to 0} \frac{1}{\sqrt{1+x} + 1}. \] Substituting \( x = 0 \): \[ \frac{1}{2} = f(0). \]

Answer 2

Given:
\[ f(x) = \frac{\sqrt{1+x} - 1}{x} \] and \( f(x) \) is continuous at \( x = 0 \), find \( f(0) \).

Step 1: To ensure continuity at \( x = 0 \), define:
\[ f(0) = \lim_{x \to 0} \frac{\sqrt{1+x} - 1}{x} \]

Step 2: Evaluate the limit using the conjugate:
\[ \lim_{x \to 0} \frac{\sqrt{1+x} - 1}{x} \times \frac{\sqrt{1+x} + 1}{\sqrt{1+x} + 1} = \lim_{x \to 0} \frac{(1+x) - 1}{x(\sqrt{1+x} + 1)} = \lim_{x \to 0} \frac{x}{x(\sqrt{1+x} + 1)} = \lim_{x \to 0} \frac{1}{\sqrt{1+x} + 1} \]

Step 3: Substitute \( x = 0 \):
\[ \frac{1}{\sqrt{1+0} + 1} = \frac{1}{1 + 1} = \frac{1}{2} \]

Therefore, to make \( f(x) \) continuous at \( x = 0 \), we set:
\[ \boxed{ f(0) = \frac{1}{2} } \]