If `lim_(xtoa^(+))f(x)=l=lim_(xtoa^(-))g(x)` and `lim_(xtoa^(-))f(x)=m=lim_(xtoa^(+))g(x)`, the function `f(x).g(x)` is

Options

A. is not continuous at `x=a`
B. has a limit when `xtoa` and equal to `lm`
C. is continuous at x=a
D. has a limit when `xtoa` and is not equal to `lm`

Answer

To solve the problem, we need to analyze the limits of the functions \( f(x) \) and \( g(x) \) as \( x \) approaches \( a \) from both sides.

### Step-by-Step Solution:

1. **Understanding the Given Limits**:
We are given:
\[ \lim_{x \to a^+} f(x) = L = \lim_{x \to a^-} g(x) \]
\[ \lim_{x \to a^-} f(x) = M = \lim_{x \to a^+} g(x) \]

2. **Finding the Left-Hand Limit of \( f(x)g(x) \)**:
We can express the limit of the product \( f(x)g(x) \) as:
\[ \lim_{x \to a^-} f(x)g(x) = \lim_{x \to a^-} f(x) \cdot \lim_{x \to a^-} g(x) \]
From the given information, we know:
\[ \lim_{x \to a^-} f(x) = M \quad \text{and} \quad \lim_{x \to a^-} g(x) = L \]
Therefore:
\[ \lim_{x \to a^-} f(x)g(x) = M \cdot L \]

3. **Finding the Right-Hand Limit of \( f(x)g(x) \)**:
Similarly, we can express the limit of the product as:
\[ \lim_{x \to a^+} f(x)g(x) = \lim_{x \to a^+} f(x) \cdot \lim_{x \to a^+} g(x) \]
From the given information, we know:
\[ \lim_{x \to a^+} f(x) = L \quad \text{and} \quad \lim_{x \to a^+} g(x) = M \]
Therefore:
\[ \lim_{x \to a^+} f(x)g(x) = L \cdot M \]

4. **Equating the Two Limits**:
Since both limits must be equal for the limit of the product to exist:
\[ \lim_{x \to a^-} f(x)g(x) = \lim_{x \to a^+} f(x)g(x) \]
This gives us:
\[ M \cdot L = L \cdot M \]
Hence, both limits are equal.

5. **Conclusion**:
Since the left-hand limit and right-hand limit are equal, we conclude that:
\[ \lim_{x \to a} f(x)g(x) = ML \]
Therefore, the function \( f(x)g(x) \) is continuous at \( x = a \) and has a limit equal to \( ML \).

### Final Answer:
The function \( f(x)g(x) \) has a limit as \( x \to a \) and is equal to \( ML \).

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