Question:

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

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For limits involving square roots, multiply by the conjugate to simplify.
Updated On: Mar 24, 2025
  • \( -\frac{1}{2} \)
  • \( \frac{1}{3} \)
  • \( \frac{1}{2} \)
  • \( -\frac{1}{3} \)
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The Correct Option is C

Solution and Explanation

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 \( x = 0 \): \[ \frac{1}{2} = f(0). \]

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