`lim_(x to 0) ((2^x +3^x + 6^x )/3)^(3//x)` is equal to :

Explanation

To solve the limit \( \lim_{x \to 0} \left(\frac{2^x + 3^x + 6^x}{3}\right)^{\frac{3}{x}} \), we can follow these steps:

### Step 1: Rewrite the limit expression
We start with the limit expression:
\[ L = \lim_{x \to 0} \left(\frac{2^x + 3^x + 6^x}{3}\right)^{\frac{3}{x}} \]

### Step 2: Identify the form of the limit
As \( x \to 0 \), \( 2^x \to 1 \), \( 3^x \to 1 \), and \( 6^x \to 1 \). Therefore, the expression inside the limit approaches:
\[ \frac{1 + 1 + 1}{3} = 1 \]
Thus, we have the form \( 1^{\infty} \).

### Step 3: Use the exponential limit property
For limits of the form \( 1^{\infty} \), we can rewrite it using the exponential function:
\[ L = e^{\lim_{x \to 0} \left(\frac{2^x + 3^x + 6^x}{3} - 1\right) \cdot \frac{3}{x}} \]

### Step 4: Simplify the expression inside the limit
We need to simplify \( \frac{2^x + 3^x + 6^x}{3} - 1 \):
\[ \frac{2^x + 3^x + 6^x - 3}{3} \]
As \( x \to 0 \), we can use the Taylor expansion for \( a^x \) around \( x = 0 \):
\[ a^x \approx 1 + x \ln a \]
Thus,
\[ 2^x \approx 1 + x \ln 2, \quad 3^x \approx 1 + x \ln 3, \quad 6^x \approx 1 + x \ln 6 \]
Adding these up:
\[ 2^x + 3^x + 6^x \approx 3 + x (\ln 2 + \ln 3 + \ln 6) \]
So,
\[ \frac{2^x + 3^x + 6^x - 3}{3} \approx \frac{x (\ln 2 + \ln 3 + \ln 6)}{3} \]

### Step 5: Substitute back into the limit
Now substituting back, we have:
\[ L = e^{\lim_{x \to 0} \frac{x (\ln 2 + \ln 3 + \ln 6)}{3} \cdot \frac{3}{x}} = e^{\ln 2 + \ln 3 + \ln 6} \]

### Step 6: Simplify the exponent
Using the property of logarithms:
\[ \ln 2 + \ln 3 + \ln 6 = \ln(2 \cdot 3 \cdot 6) = \ln 36 \]
Thus,
\[ L = e^{\ln 36} = 36 \]

### Final Answer
The limit is:
\[ \boxed{36} \]