`Lt_(xto0)(sqrt(1+x)-1)/(x)` is equal to
(i) 0
(ii) 1
(iii) `1/2`
(iv) 2

Options

A. 0
B. 1
C. `1/2`
D. 2

Answer

To solve the limit \( \lim_{x \to 0} \frac{\sqrt{1+x} - 1}{x} \), we will follow these steps:

### Step 1: Rationalize the Numerator
We start by multiplying the numerator and the denominator by the conjugate of the numerator, which is \( \sqrt{1+x} + 1 \).

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

### Step 2: Simplify the Expression
This gives us:

\[ \lim_{x \to 0} \frac{(\sqrt{1+x})^2 - 1^2}{x(\sqrt{1+x} + 1)} \]

Using the identity \( a^2 - b^2 = (a-b)(a+b) \), we can simplify the numerator:

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

### Step 3: Cancel the Common Terms
Now we can cancel \( x \) in the numerator and denominator:

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

### Step 4: Substitute \( x = 0 \)
Now we can substitute \( x = 0 \) into the expression:

\[ \frac{1}{\sqrt{1+0} + 1} = \frac{1}{\sqrt{1} + 1} = \frac{1}{1 + 1} = \frac{1}{2} \]

### Final Answer
Thus, the limit is:

\[ \lim_{x \to 0} \frac{\sqrt{1+x} - 1}{x} = \frac{1}{2} \]

### Conclusion
The answer is option (iii) \( \frac{1}{2} \).

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