Evaluate the following limits :
`Lim_(x to 0) (3^(x)-1)/(sqrt(1+x)-1)`

Answer

To evaluate the limit

\[ L = \lim_{x \to 0} \frac{3^x - 1}{\sqrt{1+x} - 1}, \]

we first substitute \(x = 0\):

\[ L = \frac{3^0 - 1}{\sqrt{1+0} - 1} = \frac{1 - 1}{1 - 1} = \frac{0}{0}. \]

Since we have an indeterminate form \( \frac{0}{0} \), we can apply L'Hôpital's Rule, which states that if the limit results in \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \), we can take the derivative of the numerator and the denominator.

### Step 1: Differentiate the numerator and denominator

The numerator is \(f(x) = 3^x - 1\). The derivative is:

\[ f'(x) = 3^x \ln(3). \]

The denominator is \(g(x) = \sqrt{1+x} - 1\). The derivative is:

\[ g'(x) = \frac{1}{2\sqrt{1+x}}. \]

### Step 2: Apply L'Hôpital's Rule

Now we apply L'Hôpital's Rule:

\[ L = \lim_{x \to 0} \frac{f'(x)}{g'(x)} = \lim_{x \to 0} \frac{3^x \ln(3)}{\frac{1}{2\sqrt{1+x}}}. \]

### Step 3: Simplify the expression

This simplifies to:

\[ L = \lim_{x \to 0} 3^x \ln(3) \cdot 2\sqrt{1+x} = \lim_{x \to 0} 2 \ln(3) \cdot 3^x \cdot \sqrt{1+x}. \]

### Step 4: Evaluate the limit

Now, substituting \(x = 0\):

\[ L = 2 \ln(3) \cdot 3^0 \cdot \sqrt{1+0} = 2 \ln(3) \cdot 1 \cdot 1 = 2 \ln(3). \]

Thus, the final answer is:

\[ \boxed{2 \ln(3)}. \]