Question

Given a real-valued function \( f \) such that: \[ f(x) = \begin{cases} \frac{\tan^2\{x\}}{x^2 - \lfloor x \rfloor^2}, & \text{for } x > 0 \\ 1, & \text{for } x = 0 \\ \sqrt{\{x\} \cot\{x\}}, & \text{for } x < 0 \end{cases} \] Then:

• A. \({LHL} = 1\)
• B. \({RHL} = \sqrt{\cot 1}\)
• C. \(\lim\limits_{x \to 0} f(x) { exists}\)
• D. \(\lim\limits_{x \to 0} f(x) { does not exist}\)

Hint:

For piecewise functions, always check left-hand and right-hand limits separately.

Answer 1

The Correct Option is D

Right-hand limit (RHL): Approaching \( x \to 0^+ \), we substitute in the first case: \[ \lim\limits_{h \to 0^+} \frac{\tan^2 h}{h^2} = \lim\limits_{h \to 0^+} \frac{\tan^2 h}{h^2} = 1 \] 
- Left-hand limit (LHL): Approaching \( x \to 0^- \), we substitute in the third case: \[ \lim\limits_{h \to 0^-} \sqrt{(-h) \cot(-h)} \] Using \( \cot(-h) = -\cot h \), we get: \[ \lim\limits_{h \to 0^-} \sqrt{(1-h) \cot(1-h)} \] which is not equal to \( \lim\limits_{x \to 0^+} f(x) \), thus: \[ \lim\limits_{x \to 0} f(x) { does not exist}. \]