`A_(i)=(x-a_(i))/(|x-a_(i)|),i=1,2,...,n," and "a_(1)lta_(2)lta_(3)lt...lta_(n).`
If `1lemlen,minN,` then `lim_(xtoa_(m)) (A_(1)A_(2)...A_(n))`

Explanation

To solve the limit problem given in the question, we will follow these steps:

### Step 1: Understand the Given Expression
We have:
\[ A_i = \frac{x - a_i}{|x - a_i|}, \quad i = 1, 2, \ldots, n \]
where \( a_1 < a_2 < \ldots < a_n \) and \( 1 \leq m \leq n \).

### Step 2: Define the Limit
We need to find:
\[ \lim_{x \to a_m} (A_1 A_2 \ldots A_n) \]

### Step 3: Analyze the Behavior of \( A_i \)
The expression \( A_i \) behaves differently depending on whether \( x \) approaches \( a_m \) from the left (denoted \( a_m^- \)) or from the right (denoted \( a_m^+ \)).

- **For \( x \to a_m^+ \)** (approaching from the right):
- If \( x > a_i \) for \( i < m \), then \( A_i = 1 \).
- If \( x = a_m \), \( A_m = 0 \) (since \( x - a_m = 0 \)).
- If \( x < a_i \) for \( i > m \), then \( A_i = -1 \).

Thus, for \( x \to a_m^+ \):
\[ A_1 A_2 \ldots A_n = 1^{m-1} \cdot 0 \cdot (-1)^{n-m} = 0 \]

- **For \( x \to a_m^- \)** (approaching from the left):
- If \( x < a_i \) for \( i \leq m \), then \( A_i = -1 \) for \( i = 1, 2, \ldots, m \).
- If \( x = a_m \), \( A_m = 0 \).
- If \( x > a_i \) for \( i > m \), then \( A_i = 1 \).

Thus, for \( x \to a_m^- \):
\[ A_1 A_2 \ldots A_n = (-1)^m \cdot 1^{n-m} = (-1)^m \]

### Step 4: Evaluate the Limits
Now we evaluate the limits:
- From the right:
\[ \lim_{x \to a_m^+} (A_1 A_2 \ldots A_n) = 0 \]

- From the left:
\[ \lim_{x \to a_m^-} (A_1 A_2 \ldots A_n) = (-1)^m \]

### Step 5: Conclusion
Since the left-hand limit and the right-hand limit are not equal:
\[ \lim_{x \to a_m} (A_1 A_2 \ldots A_n) \text{ does not exist.} \]

### Final Answer
The limit does not exist.

---