Question

\[ \lim_{x\to 0}\frac{\sqrt{1+x}-1}{x}= \]

• A. \(0\)
• B. \(\frac{1}{2}\)
• C. \(1\)
• D. \(\infty\)

Hint:

For limits involving \(\sqrt{1+x}-1\), rationalize the numerator to remove the indeterminate form.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
Evaluating the limit as \( x \to 0 \) results in the indeterminate form \( 0/0 \) because \( \sqrt{1+0} - 1 = 0 \) and the denominator is 0.
To resolve this, we can use algebraic techniques like rationalization or differentiation rules like L'Hospital's rule.
Step 2: Key Formula or Approach:
Method 1: Rationalization (multiply numerator and denominator by the conjugate).
Method 2: L'Hospital's Rule (differentiate top and bottom separately).
Step 3: Detailed Explanation:

Using Rationalization:
Multiply numerator and denominator by \( (\sqrt{1+x} + 1) \):
\[ \lim_{x \to 0} \frac{(\sqrt{1+x} - 1)(\sqrt{1+x} + 1)}{x(\sqrt{1+x} + 1)} \]
Apply the identity \( (a-b)(a+b) = a^2 - b^2 \):
\[ \lim_{x \to 0} \frac{(1+x) - 1}{x(\sqrt{1+x} + 1)} = \lim_{x \to 0} \frac{x}{x(\sqrt{1+x} + 1)} \]
Cancel \( x \) from the top and bottom:
\[ \lim_{x \to 0} \frac{1}{\sqrt{1+x} + 1} \]
Now, substitute \( x = 0 \):
\[ \frac{1}{\sqrt{1+0} + 1} = \frac{1}{1 + 1} = \frac{1}{2} \]

Using L'Hospital's Rule:
Differentiate numerator: \( \frac{d}{dx}(\sqrt{1+x} - 1) = \frac{1}{2\sqrt{1+x}} \).
Differentiate denominator: \( \frac{d}{dx}(x) = 1 \).
The limit becomes:
\[ \lim_{x \to 0} \frac{1/(2\sqrt{1+x})}{1} = \frac{1}{2\sqrt{1+0}} = \frac{1}{2} \]

Step 4: Final Answer:
The value of the limit is 1/2.