If `lim_(xtoa)f(x)=1` and `lim_(xtoa)g(x)=oo` then `lim_(xtoa){f(x)}^(g(x))=e^(lim_(xtoa)(f(x)-1)g(x))` `lim_(xto0)((a^(x)+b^(x)+c^(x))/3)^(2/x)` is equal to

Options

A. `a^(2//3)+b^(2//3)+c^(2//3)`
B. `abc`
C. `(abc)^(2//3)`
D. `1`

Answer

To solve the limit problem given, we will follow the steps outlined in the video transcript. Let's break it down step by step.

### Step-by-Step Solution

1. **Identify the Limit Expression**:
We need to evaluate the limit:
\[ \lim_{x \to 0} \left( \frac{a^x + b^x + c^x}{3} \right)^{\frac{2}{x}} \]

2. **Recognize the Form**:
As \( x \to 0 \), both \( a^x \), \( b^x \), and \( c^x \) approach 1. Therefore, the expression inside the limit approaches:
\[ \frac{1 + 1 + 1}{3} = 1 \]
Thus, we have a \( 1^{\infty} \) form.

3. **Apply the Limit Formula**:
We can use the formula:
\[ \lim_{x \to a} f(x)^{g(x)} = e^{\lim_{x \to a} (f(x) - 1) g(x)} \]
Here, let \( f(x) = \frac{a^x + b^x + c^x}{3} \) and \( g(x) = \frac{2}{x} \).

4. **Calculate \( f(x) - 1 \)**:
We need to find \( f(x) - 1 \):
\[ f(x) - 1 = \frac{a^x + b^x + c^x}{3} - 1 = \frac{a^x + b^x + c^x - 3}{3} \]

5. **Multiply by \( g(x) \)**:
Now, we multiply \( f(x) - 1 \) by \( g(x) \):
\[ (f(x) - 1) g(x) = \left(\frac{a^x + b^x + c^x - 3}{3}\right) \cdot \frac{2}{x} = \frac{2(a^x + b^x + c^x - 3)}{3x} \]

6. **Evaluate the Limit**:
We need to evaluate:
\[ \lim_{x \to 0} \frac{2(a^x + b^x + c^x - 3)}{3x} \]
Using the Taylor expansion for \( a^x \), \( b^x \), and \( c^x \) around \( x = 0 \):
\[ a^x \approx 1 + x \ln a, \quad b^x \approx 1 + x \ln b, \quad c^x \approx 1 + x \ln c \]
Therefore:
\[ a^x + b^x + c^x \approx 3 + x(\ln a + \ln b + \ln c) \]
Thus:
\[ a^x + b^x + c^x - 3 \approx x(\ln a + \ln b + \ln c) \]

7. **Substituting Back**:
Substituting this back into our limit gives:
\[ \lim_{x \to 0} \frac{2(x(\ln a + \ln b + \ln c))}{3x} = \lim_{x \to 0} \frac{2(\ln a + \ln b + \ln c)}{3} = \frac{2(\ln a + \ln b + \ln c)}{3} \]

8. **Final Limit**:
Now, we can substitute this result back into our exponential limit:
\[ \lim_{x \to 0} f(x)^{g(x)} = e^{\frac{2(\ln a + \ln b + \ln c)}{3}} = (a \cdot b \cdot c)^{\frac{2}{3}} \]

### Final Answer:
\[ \lim_{x \to 0} \left( \frac{a^x + b^x + c^x}{3} \right)^{\frac{2}{x}} = (abc)^{\frac{2}{3}} \]