Evaluate the following limits :
`lim_(x to 0)(sqrt(1+x)-1)/x`

Answer

To evaluate the limit

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

we first observe that directly substituting \( x = 0 \) gives us the indeterminate form \( \frac{0}{0} \). Therefore, we can apply L'Hôpital's Rule, which is useful for resolving limits of the form \( \frac{0}{0} \).

### Step 1: Differentiate the numerator and denominator

According to L'Hôpital's Rule, we differentiate the numerator and the denominator separately:

- The numerator is \( \sqrt{1+x} - 1 \). The derivative of \( \sqrt{1+x} \) is given by:

\[ \frac{d}{dx}(\sqrt{1+x}) = \frac{1}{2\sqrt{1+x}}. \]

Thus, the derivative of the numerator is:

\[ \frac{1}{2\sqrt{1+x}}. \]

- The denominator is \( x \), and its derivative is simply:

\[ \frac{d}{dx}(x) = 1. \]

### Step 2: Apply L'Hôpital's Rule

Now we apply L'Hôpital's Rule:

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

### Step 3: Evaluate the limit

Now, we substitute \( x = 0 \) into the limit:

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

### Final Answer

Thus, the value of the limit is

\[ \frac{1}{2}. \]

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