I apologize if my previous explanation was not clear enough. Let me provide a more detailed step-by-step explanation to address your question.

We are asked to find the limit:
[
L = \lim_{x \to 0} \frac{x}{a} \left[ \frac{b}{x} \right]
]
where ([ \cdot ]) denotes the floor function (the greatest integer less than or equal to the argument).

Understanding the Floor Function:

The floor function (\left[ \frac{b}{x} \right]) returns the largest integer less than or equal to (\frac{b}{x}). As (x) approaches (0), (\frac{b}{x}) behaves differently depending on whether (x) approaches from the right ((x \to 0^+)) or from the left ((x \to 0^-)).

Case 1: (x \to 0^+) (Approaching from the Right)

Let’s consider (x > 0) and approaching (0) from the right.

Behavior of (\frac{b}{x}):
If (b > 0), then (\frac{b}{x} \to +\infty).
If (b < 0), then (\frac{b}{x} \to -\infty).
For simplicity, let's assume (b > 0) (the argument is similar if (b < 0) with sign changes).
Defining (n): Let’s define:
[
n = \left[ \frac{b}{x} \right]
]
So (n) is the largest integer less than or equal to (\frac{b}{x}).
Expressing (\frac{b}{x}) in Terms of (n): Since (n \leq \frac{b}{x} < n + 1), we can write:
[
\frac{b}{x} = n + \theta
]
where (\theta \in [0,1)), representing the fractional part of (\frac{b}{x}).
Solving for (n x): Multiplying both sides by (x):
[
b = n x + \theta x
]
Rearranging:
[
n x = b - \theta x
]
As (x \to 0^+), (\theta x \to 0) (since (\theta) is bounded between 0 and 1, and (x \to 0)).
Evaluating the Limit (L): The expression becomes:
[
L = \lim_{x \to 0^+} \frac{x}{a} \cdot n = \lim_{x \to 0^+} \frac{x}{a} \cdot \left( \frac{b}{x} - \theta \right)
]
Simplify:
[
L = \lim_{x \to 0^+} \left( \frac{b}{a} - \frac{x}{a} \theta \right)
]
Since (\frac{x}{a} \theta \to 0) as (x \to 0^+):
[
L = \frac{b}{a}
]
Case 2: (x \to 0^-) (Approaching from the Left)

Now, consider (x < 0) and approaching (0) from the left.

Behavior of (\frac{b}{x}):
If (b > 0), then (\frac{b}{x} \to -\infty).
If (b < 0), then (\frac{b}{x} \to +\infty).
Again, we'll assume (b > 0) for this explanation.
Defining (n): Let’s define:
[
n = \left[ \frac{b}{x} \right]
]
Since (\frac{b}{x}) is a large negative number, (n) is a large negative integer.
Expressing (\frac{b}{x}) in Terms of (n): We have:
[
n \leq \frac{b}{x} < n + 1
]
Multiplying all parts by (x < 0) (remember, multiplying an inequality by a negative number reverses the inequality signs):
[
n x \geq b > (n + 1) x
]
So:
[
b = n x + \theta x
]
where (\theta = \frac{b}{x} - n), and (\theta \in [0,1)).
Solving for (n x): Rearranging:
[
n x = b - \theta x
]
As (x \to 0^-), (\theta x \to 0).
Evaluating the Limit (L): The expression becomes:
[
L = \lim_{x \to 0^-} \frac{x}{a} \cdot n = \lim_{x \to 0^-} \frac{x}{a} \cdot \left( \frac{b}{x} - \theta \right)
]
Simplify:
[
L = \lim_{x \to 0^-} \left( \frac{b}{a} - \frac{x}{a} \theta \right)
]
Since (\frac{x}{a} \theta \to 0) as (x \to 0^-):
[
L = \frac{b}{a}
]
Conclusion:

In both cases, whether (x) approaches (0) from the positive or negative side, the limit of the expression is (\frac{b}{a}).

Answer:

[
\lim_{x \to 0} \frac{x}{a} \left[ \frac{b}{x} \right] = \boxed{\dfrac{b}{a}}
]

To directly address your question:

When ( n = \frac{b}{x} )?

The value ( n ) is defined as ( n = \left[ \frac{b}{x} \right] ), which means it is the integer part of ( \frac{b}{x} ). It is not exactly equal to ( \frac{b}{x} ), but rather the greatest integer less than or equal to ( \frac{b}{x} ).

By expressing ( \frac{b}{x} ) as ( n + \theta ) with ( \theta \in [0,1) ), we account for the fractional part. This allows us to write ( n = \frac{b}{x} - \theta ). As ( x \to 0 ), ( \theta ) remains bounded between 0 and 1, while ( \frac{b}{x} ) becomes unbounded (approaches infinity or negative infinity).

So, ( n ) is approximately ( \frac{b}{x} ) minus a small fractional part, but it is not exactly equal to ( \frac{b}{x} ). In the limit process, this distinction becomes insignificant because the term involving ( \theta ) vanishes as ( x \to 0 ).