`lim_(x to 0) ((4^(x)+9^(x))/(2))^(1//x)` equals:

Options

A. 2
B. 6
C. 16
D. 112

Explanation

To solve the limit problem \( \lim_{x \to 0} \left( \frac{4^x + 9^x}{2} \right)^{\frac{1}{x}} \), we will follow these steps:

### Step 1: Identify the Indeterminate Form
First, we substitute \( x = 0 \):
\[ \frac{4^0 + 9^0}{2} = \frac{1 + 1}{2} = 1 \]
Thus, we have:
\[ \left(1\right)^{\frac{1}{0}} = 1^{\infty} \]
This is an indeterminate form, so we need to apply logarithms to resolve it.

### Step 2: Take the Natural Logarithm
Let \( L = \lim_{x \to 0} \left( \frac{4^x + 9^x}{2} \right)^{\frac{1}{x}} \). Taking the natural logarithm on both sides gives:
\[ \ln L = \lim_{x \to 0} \frac{1}{x} \ln\left( \frac{4^x + 9^x}{2} \right) \]

### Step 3: Simplify the Logarithm
We can rewrite the logarithm:
\[ \ln L = \lim_{x \to 0} \frac{1}{x} \left( \ln(4^x + 9^x) - \ln(2) \right) \]
This can be further simplified to:
\[ \ln L = \lim_{x \to 0} \frac{\ln(4^x + 9^x)}{x} - \frac{\ln(2)}{x} \]
As \( x \to 0 \), \( \frac{\ln(2)}{x} \) approaches infinity, so we need to focus on the first term.

### Step 4: Apply L'Hôpital's Rule
The limit \( \lim_{x \to 0} \frac{\ln(4^x + 9^x)}{x} \) is of the form \( \frac{0}{0} \), so we can apply L'Hôpital's Rule:
\[ \ln(4^x + 9^x) \text{ differentiates to } \frac{4^x \ln(4) + 9^x \ln(9)}{4^x + 9^x} \]
Thus, we have:
\[ \ln L = \lim_{x \to 0} \frac{4^x \ln(4) + 9^x \ln(9)}{4^x + 9^x} \]

### Step 5: Evaluate the Limit
As \( x \to 0 \):
\[ 4^x \to 1 \quad \text{and} \quad 9^x \to 1 \]
So:
\[ \ln L = \frac{1 \cdot \ln(4) + 1 \cdot \ln(9)}{1 + 1} = \frac{\ln(4) + \ln(9)}{2} \]
Using properties of logarithms, we can combine:
\[ \ln L = \frac{\ln(4 \cdot 9)}{2} = \frac{\ln(36)}{2} \]

### Step 6: Solve for \( L \)
Exponentiating both sides gives:
\[ L = e^{\frac{\ln(36)}{2}} = 36^{\frac{1}{2}} = 6 \]

### Conclusion
Thus, the limit is:
\[ \lim_{x \to 0} \left( \frac{4^x + 9^x}{2} \right)^{\frac{1}{x}} = 6 \]