The value of `lim_(xrarr0) {(a^x+b^x+c^x)/(3)}^(1//x)`, is

Options

A. abc
B. `(abc)^1//3`
C. `(1)/(3)abc `
D. none of these

Explanation

To solve the limit \( \lim_{x \to 0} \left( \frac{a^x + b^x + c^x}{3} \right)^{\frac{1}{x}} \), we can follow these steps:

### Step 1: Identify the limit form
First, we substitute \( x = 0 \) into the expression:
\[ \frac{a^0 + b^0 + c^0}{3} = \frac{1 + 1 + 1}{3} = 1 \]
Thus, we have:
\[ \left( \frac{a^x + b^x + c^x}{3} \right)^{\frac{1}{x}} \to 1^{\infty} \text{ as } x \to 0 \]
This is an indeterminate form of type \( 1^{\infty} \).

### Step 2: Apply the logarithm
To resolve this indeterminate form, we can take the natural logarithm:
\[ L = \lim_{x \to 0} \frac{1}{x} \ln \left( \frac{a^x + b^x + c^x}{3} \right) \]

### Step 3: Simplify the logarithm
Now, we can write:
\[ L = \lim_{x \to 0} \frac{1}{x} \left( \ln(a^x + b^x + c^x) - \ln(3) \right) \]
Since \( \ln(3) \) is a constant, we can focus on \( \ln(a^x + b^x + c^x) \).

### Step 4: Evaluate \( a^x + b^x + c^x \)
As \( x \to 0 \):
\[ a^x \to 1, \quad b^x \to 1, \quad c^x \to 1 \]
Thus:
\[ a^x + b^x + c^x \to 3 \]
This means:
\[ \ln(a^x + b^x + c^x) \to \ln(3) \]

### Step 5: Apply L'Hôpital's Rule
Now we have:
\[ L = \lim_{x \to 0} \frac{\ln(a^x + b^x + c^x) - \ln(3)}{x} \]
This is a \( 0/0 \) form, so we can apply L'Hôpital's Rule:
\[ L = \lim_{x \to 0} \frac{\frac{d}{dx} \left( \ln(a^x + b^x + c^x) \right)}{\frac{d}{dx}(x)} \]
The derivative of \( x \) is \( 1 \).

### Step 6: Differentiate the numerator
Using the chain rule:
\[ \frac{d}{dx} \left( \ln(a^x + b^x + c^x) \right) = \frac{1}{a^x + b^x + c^x} \cdot (a^x \ln a + b^x \ln b + c^x \ln c) \]
Thus:
\[ L = \lim_{x \to 0} \frac{a^x \ln a + b^x \ln b + c^x \ln c}{a^x + b^x + c^x} \]

### Step 7: Substitute \( x = 0 \)
Now substituting \( x = 0 \):
\[ L = \frac{1 \cdot \ln a + 1 \cdot \ln b + 1 \cdot \ln c}{3} = \frac{\ln a + \ln b + \ln c}{3} \]

### Step 8: Exponentiate to find the limit
Thus, we find:
\[ \lim_{x \to 0} \left( \frac{a^x + b^x + c^x}{3} \right)^{\frac{1}{x}} = e^L = e^{\frac{\ln a + \ln b + \ln c}{3}} = (abc)^{\frac{1}{3}} \]

### Final Answer
The value of the limit is:
\[ \boxed{(abc)^{\frac{1}{3}}} \]