Use the formula `Lim_(xto 0) (a^(x)-1)/x = log_(e)a " to find " Lim_(xto0) (2^(x)-1)/((1+x)^(1//2)-1)`

Answer

To solve the limit problem \( \lim_{x \to 0} \frac{2^x - 1}{\sqrt{1+x} - 1} \) using the formula \( \lim_{x \to 0} \frac{a^x - 1}{x} = \log_e a \), we will follow these steps:

### Step 1: Identify the limit expression
We need to evaluate:
\[ \lim_{x \to 0} \frac{2^x - 1}{\sqrt{1+x} - 1} \]

### Step 2: Apply the formula for the numerator
Using the formula, we know:
\[ \lim_{x \to 0} \frac{2^x - 1}{x} = \log_e 2 \]
Thus, we can rewrite our limit as:
\[ \lim_{x \to 0} \frac{2^x - 1}{\sqrt{1+x} - 1} = \lim_{x \to 0} \frac{2^x - 1}{x} \cdot \frac{x}{\sqrt{1+x} - 1} \]

### Step 3: Simplify the denominator
We need to simplify the expression \( \sqrt{1+x} - 1 \). We can multiply and divide by the conjugate:
\[ \sqrt{1+x} - 1 = \frac{(\sqrt{1+x} - 1)(\sqrt{1+x} + 1)}{\sqrt{1+x} + 1} = \frac{1+x - 1}{\sqrt{1+x} + 1} = \frac{x}{\sqrt{1+x} + 1} \]

### Step 4: Substitute back into the limit
Now we can substitute this back into our limit:
\[ \lim_{x \to 0} \frac{2^x - 1}{\sqrt{1+x} - 1} = \lim_{x \to 0} \frac{2^x - 1}{x} \cdot \frac{x}{\frac{x}{\sqrt{1+x} + 1}} = \lim_{x \to 0} \frac{2^x - 1}{x} \cdot (\sqrt{1+x} + 1) \]

### Step 5: Evaluate the limit
Now we can evaluate the limit:
1. From the earlier step, we know:
\[ \lim_{x \to 0} \frac{2^x - 1}{x} = \log_e 2 \]
2. For \( \sqrt{1+x} + 1 \):
\[ \lim_{x \to 0} (\sqrt{1+x} + 1) = \sqrt{1+0} + 1 = 1 + 1 = 2 \]

### Step 6: Combine the results
Now we combine the results:
\[ \lim_{x \to 0} \frac{2^x - 1}{\sqrt{1+x} - 1} = \log_e 2 \cdot 2 = 2 \log_e 2 \]

### Final Answer
Thus, the final answer is:
\[ \lim_{x \to 0} \frac{2^x - 1}{\sqrt{1+x} - 1} = 2 \log_e 2 \]

---