If \( f(x) = \begin{cases} \frac{\sin |x|}{x}, & \text{for } [x] \ne 0 \\ 0, & \text{for } [x] = 0 \end{cases} \) where, \([x]\) denotes the greatest integer less than or equal to \(x\), then \(\lim_{x \to 0} f(x)\) is equal to

Question

Let the function \( g: (-\infty, 0) \rightarrow \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \) be given by \( g(u) = 2 \tan^{-1}(e^u) - \frac{\pi}{2} \). Determine the properties of \( g \).

Answer 1

To solve this problem, we need to evaluate the limit:

\(\lim_{x \to 0} f(x)\)

where the function \( f(x) \) is defined as:

\( f(x) = \begin{cases} \frac{\sin |x|}{x}, & \text{for } [x] \ne 0 \\ 0, & \text{for } [x] = 0 \end{cases} \)

The key to solving this problem is understanding how the greatest integer function (floor function) \([x]\) behaves as \(x\) approaches 0. The floor function \([x]\) gives the greatest integer less than or equal to \(x\).

Step-by-Step Evaluation:

As \(x\) approaches 0, depending on the direction from which \(x\) approaches, \([x]\) can be -1 or 0.
For small positive values of \(x\) (approaching 0 from the right), \([x] = 0\).
For small negative values of \(x\) (approaching 0 from the left), \([x] = -1\).

Therefore, let us consider the left-hand and right-hand limits:

Left-hand limit (\(x \to 0^-\)): In this case, \([x] = -1\). Thus, \(f(x) = \frac{\sin |x|}{x}.\)
Right-hand limit (\(x \to 0^+\)): In this case, \([x] = 0\). Therefore, \(f(x) = 0\).

If the left-hand limit and the right-hand limit are not equal, then the limit does not exist:

\(\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{\sin |x|}{x}\) yields an expression that generally does not simplify to a standard limit formula.
\(\lim_{x \to 0^+} f(x) = 0\) because \([x] = 0\) implies \(f(x) = 0\).

Because \(\lim_{x \to 0^-} f(x) \ne \lim_{x \to 0^+} f(x)\), the limit does not exist.

Conclusion:

The limit \(\lim_{x \to 0} f(x)\) does not exist. Therefore, the correct answer is "Does not exist."

Answer 2