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

Concept: When a limit contains a square root expression, rationalization is often useful.

Step 1: Given limit is: \[ \lim_{x\to 0}\frac{\sqrt{1+x}-1}{x} \]

Step 2: Rationalize the numerator. Multiply numerator and denominator by: \[ \sqrt{1+x}+1 \] \[ \frac{\sqrt{1+x}-1}{x} \times \frac{\sqrt{1+x}+1}{\sqrt{1+x}+1} \]

Step 3: Apply identity \((a-b)(a+b)=a^2-b^2\). \[ = \frac{(1+x)-1}{x(\sqrt{1+x}+1)} \] \[ = \frac{x}{x(\sqrt{1+x}+1)} \] Cancel \(x\): \[ = \frac{1}{\sqrt{1+x}+1} \]

Step 4: Now put \(x=0\). \[ \frac{1}{\sqrt{1+0}+1} = \frac{1}{1+1} = \frac{1}{2} \] Therefore, \[ \boxed{\frac{1}{2}} \]