Let` a_(1)gta_(2)gta_(3)gt...gta_(n)gt1.`
`p_(1)gtp_(2)gtp_(3)gt...gtp_(n)gt0" such that "p_(1)+p_(2)+p_(3)+...+p_(n)=1.`
Also, `F(x)=(p_(1)a_(1)^(x)+p_(n)a_(n)^(x))^(1//x)`.
`lim_(xto0^(+)) F(x)" equals "`

Options

A. `{:(a,b,c,d),(s,r,q,p):}`
B. `{:(a,b,c,d),(q,p,s,p):}`
C. `{:(a,b,c,d),(s,r,p,q):}`
D. `{:(a,b,c,d),(p,p,q,r):}`

Explanation

To solve the limit problem given in the question, we need to analyze the function \( F(x) \) defined as:

\[ F(x) = (p_1 a_1^x + p_n a_n^x)^{\frac{1}{x}} \]

We are tasked with finding:

\[ \lim_{x \to 0^+} F(x) \]

### Step 1: Rewrite the Function

First, we can rewrite \( F(x) \) in a more manageable form. Recognizing that as \( x \to 0 \), both \( a_1^x \) and \( a_n^x \) approach 1 (since any number raised to the power of 0 is 1), we can express:

\[ F(x) = (p_1 a_1^x + p_n a_n^x)^{\frac{1}{x}} = (p_1 \cdot 1 + p_n \cdot 1)^{\frac{1}{x}} = (p_1 + p_n)^{\frac{1}{x}} \]

### Step 2: Evaluate the Limit

Next, we need to evaluate the limit:

\[ \lim_{x \to 0^+} F(x) = \lim_{x \to 0^+} (p_1 a_1^x + p_n a_n^x)^{\frac{1}{x}} \]

Using the property of limits, we can take the logarithm of \( F(x) \):

\[ \ln F(x) = \frac{1}{x} \ln(p_1 a_1^x + p_n a_n^x) \]

### Step 3: Apply L'Hôpital's Rule

As \( x \to 0^+ \), both the numerator and denominator approach 0, so we can apply L'Hôpital's Rule:

\[ \lim_{x \to 0^+} \frac{\ln(p_1 a_1^x + p_n a_n^x)}{x} \]

Differentiating the numerator and denominator:

1. The derivative of the numerator \( \ln(p_1 a_1^x + p_n a_n^x) \) is:

\[ \frac{d}{dx} \ln(p_1 a_1^x + p_n a_n^x) = \frac{p_1 a_1^x \ln a_1 + p_n a_n^x \ln a_n}{p_1 a_1^x + p_n a_n^x} \]

2. The derivative of the denominator \( x \) is simply 1.

Thus, we have:

\[ \lim_{x \to 0^+} \frac{p_1 a_1^x \ln a_1 + p_n a_n^x \ln a_n}{p_1 a_1^x + p_n a_n^x} \]

### Step 4: Evaluate the Limit

As \( x \to 0^+ \), \( a_1^x \) and \( a_n^x \) approach 1, so we can substitute:

\[ \lim_{x \to 0^+} \frac{p_1 \cdot 1 \cdot \ln a_1 + p_n \cdot 1 \cdot \ln a_n}{p_1 \cdot 1 + p_n \cdot 1} = \frac{p_1 \ln a_1 + p_n \ln a_n}{p_1 + p_n} \]

### Step 5: Exponentiate the Result

Finally, we exponentiate the result to find \( F(0) \):

\[ F(0) = e^{\frac{p_1 \ln a_1 + p_n \ln a_n}{p_1 + p_n}} = (a_1^{p_1} a_n^{p_n})^{\frac{1}{p_1 + p_n}} = a_1^{\frac{p_1}{p_1 + p_n}} a_n^{\frac{p_n}{p_1 + p_n}} \]

### Final Answer

Thus, we conclude that:

\[ \lim_{x \to 0^+} F(x) = a_1^{\frac{p_1}{p_1 + p_n}} a_n^{\frac{p_n}{p_1 + p_n}} \]