Evaluate `lim_(x to 0) (2^(x)-1)/((1+x)^(1//2)-1)`

Options

A. `(1)/(8sqrt(3))`
B. `(1)/(4sqrt(3))`
C. 0
D. none of these

Explanation

To evaluate the limit

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

we start by substituting \(x = 0\):

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

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

### Step 1: Differentiate the numerator and denominator

**Numerator:**

The numerator is \(2^x - 1\). The derivative is:

\[ \frac{d}{dx}(2^x - 1) = 2^x \ln(2). \]

**Denominator:**

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

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

### Step 2: Apply L'Hôpital's Rule

Now we can rewrite the limit using the derivatives:

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

### Step 3: Substitute \(x = 0\) again

Now, substituting \(x = 0\):

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

### Step 4: Final result

Thus, the limit evaluates to:

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