`lim_(x->0)((4^x+9^x)/2)^(1/x)`

Options

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

Answer

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

### Step 1: Recognize the limit form
We start by substituting \( x = 0 \) into the expression:
\[ \frac{4^0 + 9^0}{2} = \frac{1 + 1}{2} = 1 \]
Thus, we have the limit in the form \( 1^{\infty} \), which is an indeterminate form.

**Hint:** When you encounter \( 1^{\infty} \), consider using the exponential limit transformation.

### Step 2: Use the exponential limit transformation
We can rewrite the limit using the exponential function:
\[ \lim_{x \to 0} \left( \frac{4^x + 9^x}{2} \right)^{\frac{1}{x}} = e^{\lim_{x \to 0} \left( \frac{4^x + 9^x}{2} - 1 \right) \cdot \frac{1}{x}} \]

**Hint:** This transformation helps us to convert the limit into a more manageable form.

### Step 3: Simplify the expression inside the limit
Now, we need to compute:
\[ \lim_{x \to 0} \left( \frac{4^x + 9^x}{2} - 1 \right) \cdot \frac{1}{x} \]
Using the Taylor expansion for \( a^x \) around \( x = 0 \):
\[ a^x \approx 1 + x \ln a \quad \text{for small } x \]
we can expand \( 4^x \) and \( 9^x \):
\[ 4^x \approx 1 + x \ln 4, \quad 9^x \approx 1 + x \ln 9 \]
Thus,
\[ \frac{4^x + 9^x}{2} \approx \frac{(1 + x \ln 4) + (1 + x \ln 9)}{2} = 1 + \frac{x (\ln 4 + \ln 9)}{2} \]

**Hint:** Use Taylor series expansions for exponential functions to simplify the limit.

### Step 4: Substitute back into the limit
Now substituting back, we have:
\[ \frac{4^x + 9^x}{2} - 1 \approx \frac{x (\ln 4 + \ln 9)}{2} \]
Thus, our limit becomes:
\[ \lim_{x \to 0} \left( \frac{x (\ln 4 + \ln 9)}{2} \right) \cdot \frac{1}{x} = \lim_{x \to 0} \frac{\ln 4 + \ln 9}{2} = \frac{\ln 4 + \ln 9}{2} \]

**Hint:** The \( x \) terms will cancel out, leaving a constant.

### Step 5: Calculate the final limit
Now we can simplify:
\[ \frac{\ln 4 + \ln 9}{2} = \frac{\ln(4 \cdot 9)}{2} = \frac{\ln 36}{2} = \ln 6 \]
Thus, we have:
\[ \lim_{x \to 0} \left( \frac{4^x + 9^x}{2} \right)^{\frac{1}{x}} = e^{\ln 6} = 6 \]

**Final Answer:**
\[ \lim_{x \to 0} \left( \frac{4^x + 9^x}{2} \right)^{\frac{1}{x}} = 6 \]