Step 1: Understanding the Concept:
The problem involves finding the limit of a sum of floor functions. We use the property that for \(x \to 0\), \(\frac{\sin kx}{kx}<1\). Therefore, \(k \frac{\sin kx}{x} = k^{2} \frac{\sin kx}{kx}\) will be slightly less than \(k^{2}\).
Step 2: Key Formula or Approach:
1. \(\lim_{x\to 0} \frac{\sin \theta}{\theta} = 1\), but \(\frac{\sin \theta}{\theta}<1\) for \(\theta \neq 0\).
2. For any integer \(N\), if \(N - 1<y<N\), then \([y] = N - 1\).
: Detailed Explanation:
Let's analyze the general term \([ \frac{k \sin kx}{x} ]\):
\[ \frac{k \sin kx}{x} = k^{2} \left( \frac{\sin kx}{kx} \right) \]
As \(x \to 0\), \(\frac{\sin kx}{kx} \to 1^{-}\) (approaches 1 from below).
So, \(k^{2} \left( \frac{\sin kx}{kx} \right) \to (k^{2})^{-}\).
This means the value is strictly between \(k^{2} - 1\) and \(k^{2}\).
Thus, \([ \frac{k \sin kx}{x} ] = k^{2} - 1\).
The function \(f(x)\) is the sum of such terms for \(k = 1\) to \(10\):
\[ \lim_{x\to 0} f(x) = \sum_{k=1}^{10} (k^{2} - 1) \]
\[ = \left( \sum_{k=1}^{10} k^{2} \right) - \left( \sum_{k=1}^{10} 1 \right) \]
Using the formula for sum of squares \(\frac{n(n+1)(2n+1)}{6}\):
\[ \text{Sum} = \frac{10 \times 11 \times 21}{6} - 10 \]
\[ = 385 - 10 = 375 \]
Step 3: Final Answer:
The limit is 375.