`lim_(xrarr0)(pi^(x)-1)/(sqrt(1+x)-1)`

Options

A. does not exist
B. equals `log_(e)(pi^(2))`
C. equals 1
D. lies between 10 and 11

Answer

To solve the limit \( \lim_{x \to 0} \frac{\pi^x - 1}{\sqrt{1+x} - 1} \), we can follow these steps:

### Step 1: Identify the form of the limit
First, we substitute \( x = 0 \) into the expression:
\[ \frac{\pi^0 - 1}{\sqrt{1+0} - 1} = \frac{1 - 1}{1 - 1} = \frac{0}{0} \]
This is an indeterminate form \( \frac{0}{0} \), so we can apply L'Hôpital's Rule.

**Hint:** Check if substituting the limit gives an indeterminate form like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \) before applying L'Hôpital's Rule.

### Step 2: Apply L'Hôpital's Rule
According to L'Hôpital's Rule, we differentiate the numerator and the denominator separately:
\[ \text{Numerator: } \frac{d}{dx}(\pi^x - 1) = \pi^x \ln(\pi) \]
\[ \text{Denominator: } \frac{d}{dx}(\sqrt{1+x} - 1) = \frac{1}{2\sqrt{1+x}} \]
Now we can rewrite the limit:
\[ \lim_{x \to 0} \frac{\pi^x \ln(\pi)}{\frac{1}{2\sqrt{1+x}}} \]

**Hint:** Differentiate the numerator and denominator separately when applying L'Hôpital's Rule.

### Step 3: Simplify the expression
The limit now becomes:
\[ \lim_{x \to 0} \frac{\pi^x \ln(\pi) \cdot 2\sqrt{1+x}}{1} \]
We can evaluate this limit by substituting \( x = 0 \):
\[ = \pi^0 \ln(\pi) \cdot 2\sqrt{1+0} = 1 \cdot \ln(\pi) \cdot 2 \cdot 1 = 2 \ln(\pi) \]

**Hint:** Substitute \( x = 0 \) into the simplified expression to find the limit.

### Step 4: Final answer
Thus, the limit is:
\[ \lim_{x \to 0} \frac{\pi^x - 1}{\sqrt{1+x} - 1} = 2 \ln(\pi) \]

### Summary
The final answer is:
\[ \boxed{2 \ln(\pi)} \]