Question

The value of $\lim_{x\rightarrow 0}\frac{\sqrt{1+x} - 1}{x} = $

• A. 0
• B. $1/2$
• C. 1
• D. $\infty$

Hint:

For $\sqrt{1+x}$, the binomial approximation for small $x$ is $1 + \frac{1}{2}x$. Substituting this makes the limit $(1 + \frac{1}{2}x - 1)/x = 1/2$ instantly.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The limit presents a \( 0/0 \) indeterminate form. To solve this, we can rationalize the numerator or use L'Hôpital's Rule.
Step 2: Key Formula or Approach:
1. Rationalization: Multiply the numerator and denominator by the conjugate \(\sqrt{1+x} + 1\).
2. L'Hôpital's Rule: \(\lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)}\).
Step 3: Detailed Explanation:
Using the rationalization method: \[ \lim_{x \to 0} \frac{\sqrt{1+x} - 1}{x} \times \frac{\sqrt{1+x} + 1}{\sqrt{1+x} + 1} \] \[ = \lim_{x \to 0} \frac{(1+x) - 1}{x(\sqrt{1+x} + 1)} \] \[ = \lim_{x \to 0} \frac{x}{x(\sqrt{1+x} + 1)} \] Cancel \(x\) from the numerator and denominator: \[ = \lim_{x \to 0} \frac{1}{\sqrt{1+x} + 1} \] Substitute \(x = 0\): \[ = \frac{1}{\sqrt{1+0} + 1} = \frac{1}{1 + 1} = \frac{1}{2} \]
Step 4: Final Answer:
The value of the limit is \( 1/2 \).