Question

For any real number \( y \), let \( \lfloor y \rfloor \) be the greatest integer less than or equal to \( y \) and let \( \{ y \} = y - \lfloor y \rfloor \). For \( n = 1, 2, \dots \), and for \( x \in \mathbb{R} \), let

\[ f_{2n}(x) = \begin{cases} \frac{\sin x}{x}, & x \neq 0 \\ 1, & x = 0 \end{cases} \quad \text{and} \quad f_{2n-1}(x) = \begin{cases} \frac{\sin x}{x}, & x \neq 0 \\ 1, & x = 0 \end{cases} \]

Then

\[ \lim_{x \to 0} \sum_{k=1}^{100} f_k(x) \text{ equals ............}. \]

Hint:

For sums of functions that approach a constant as \( x \to 0 \), the sum of the limits is simply the number of terms multiplied by the limit of each term.

Answer 1

Correct Answer: 50

Step 1: Understand the behavior of \( f_k(x) \).
For \( k \) even, \( f_k(x) = \frac{\sin x}{x} \) when \( x \neq 0 \) and equals 1 when \( x = 0 \). Similarly, for \( k \) odd, \( f_k(x) = \frac{\sin x}{x} \) when \( x \neq 0 \) and equals 1 when \( x = 0 \).
Step 2: Limit as \( x \to 0 \).
As \( x \to 0 \), \( \frac{\sin x}{x} \to 1 \). Therefore, for each \( f_k(x) \), as \( x \to 0 \), we have: \[ f_k(x) \to 1 \quad \text{for all} \ k. \] Thus, each term in the sum \( \sum_{k=1}^{100} f_k(x) \) approaches 1 as \( x \to 0 \).
Step 3: Compute the sum.
Since there are 100 terms in the sum, and each term tends to 1 as \( x \to 0 \), we have: \[ \sum_{k=1}^{100} f_k(x) \to 100 \quad \text{as} \quad x \to 0. \]
Final Answer: \[ \boxed{100}. \]