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

Answer

To evaluate the limit

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

we first observe that substituting \(x = 0\) directly into the expression gives us the indeterminate form \(\frac{0}{0}\). Therefore, we can apply L'Hôpital's Rule, which states that if we have an indeterminate form \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), we can take the derivative of the numerator and the derivative of the denominator.

### Step 1: Differentiate the numerator and denominator

1. **Numerator**: The numerator is \(\sqrt{1+x} - 1\). The derivative of \(\sqrt{1+x}\) with respect to \(x\) is:
\[ \frac{d}{dx}(\sqrt{1+x}) = \frac{1}{2\sqrt{1+x}}. \]
Therefore, the derivative of the numerator is:
\[ \frac{d}{dx}(\sqrt{1+x} - 1) = \frac{1}{2\sqrt{1+x}}. \]

2. **Denominator**: The denominator is \(\log(1+x)\). The derivative of \(\log(1+x)\) with respect to \(x\) is:
\[ \frac{d}{dx}(\log(1+x)) = \frac{1}{1+x}. \]

### Step 2: Apply L'Hôpital's Rule

Now we apply L'Hôpital's Rule:
\[ \lim_{x \to 0} \frac{\sqrt{1+x} - 1}{\log(1+x)} = \lim_{x \to 0} \frac{\frac{1}{2\sqrt{1+x}}}{\frac{1}{1+x}}. \]

### Step 3: Simplify the expression

This simplifies to:
\[ \lim_{x \to 0} \frac{1}{2\sqrt{1+x}} \cdot (1+x) = \lim_{x \to 0} \frac{1+x}{2\sqrt{1+x}}. \]

### Step 4: Substitute \(x = 0\)

Now we can substitute \(x = 0\) into the simplified expression:
\[ \frac{1+0}{2\sqrt{1+0}} = \frac{1}{2\sqrt{1}} = \frac{1}{2}. \]

### Final Result

Thus, the limit is:
\[ \lim_{x \to 0} \frac{\sqrt{1+x} - 1}{\log(1+x)} = \frac{1}{2}. \]

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