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

Options

A. 2
B. `log_(e) 2`
C. `log_(e) 2)`
D. `2 log_(e) 2`

Answer

To solve the limit \( \lim_{x \to 0} \frac{2^x - 1}{\sqrt{1 + x} - 1} \), we can use two methods: the expansion method and L'Hôpital's rule. Here, I will provide a step-by-step solution using the expansion method.

### Step-by-Step Solution:

1. **Identify the limit**:
\[ \lim_{x \to 0} \frac{2^x - 1}{\sqrt{1 + x} - 1} \]

2. **Expand \(2^x\) using Taylor series**:
The Taylor series expansion of \(2^x\) around \(x = 0\) is:
\[ 2^x = 1 + x \ln(2) + \frac{x^2 (\ln(2))^2}{2!} + \cdots \]
Thus,
\[ 2^x - 1 = x \ln(2) + \frac{x^2 (\ln(2))^2}{2} + \cdots \]

3. **Expand \(\sqrt{1 + x}\) using Taylor series**:
The Taylor series expansion of \(\sqrt{1 + x}\) around \(x = 0\) is:
\[ \sqrt{1 + x} = 1 + \frac{x}{2} - \frac{x^2}{8} + \cdots \]
Thus,
\[ \sqrt{1 + x} - 1 = \frac{x}{2} - \frac{x^2}{8} + \cdots \]

4. **Substitute the expansions into the limit**:
Now substituting these expansions into the limit gives:
\[ \lim_{x \to 0} \frac{x \ln(2) + \frac{x^2 (\ln(2))^2}{2} + \cdots}{\frac{x}{2} - \frac{x^2}{8} + \cdots} \]

5. **Simplify the expression**:
As \(x \to 0\), we can ignore higher-order terms. The limit simplifies to:
\[ \lim_{x \to 0} \frac{x \ln(2)}{\frac{x}{2}} = \lim_{x \to 0} \frac{2 \ln(2)}{1} = 2 \ln(2) \]

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