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

Options

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

Answer

To solve the limit

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

we first observe that substituting \(x = 0\) directly into the expression results in 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 \(2^x - 1\). The derivative of \(2^x\) is given by the formula \(2^x \ln(2)\). Therefore, the derivative of the numerator is:
\[ \frac{d}{dx}(2^x - 1) = 2^x \ln(2). \]

2. **Denominator**: The denominator is \(\sqrt{1+x} - 1\). The derivative of \(\sqrt{1+x}\) is \(\frac{1}{2\sqrt{1+x}}\). Therefore, the derivative of the denominator is:
\[ \frac{d}{dx}(\sqrt{1+x} - 1) = \frac{1}{2\sqrt{1+x}}. \]

### Step 2: Apply L'Hôpital's Rule

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

### Step 3: Simplify the Expression

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

### Step 4: Evaluate the Limit

Now we can substitute \(x = 0\):
- \(2^0 = 1\),
- \(\sqrt{1+0} = 1\).

Thus, we have:
\[ \lim_{x \to 0} 2^x \ln(2) \cdot 2\sqrt{1+x} = 1 \cdot \ln(2) \cdot 2 \cdot 1 = 2 \ln(2). \]

### Final Answer

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

---