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: \[ \lim\limits_{x \to 0} f(x) { does not exist}. \]

Answer 2

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:
Step 1: Analyze the function behavior as \( x \to 0 \)
We need to check the left-hand limit (as \( x \to 0^- \)) and right-hand limit (as \( x \to 0^+ \)) to see if the limit exists.
Right-hand limit (as \( x \to 0^+ \)):
For \( x > 0 \), the function is:
\[ f(x) = \frac{\tan^2\{x\}}{x^2 - \lfloor x \rfloor^2}. \] Here, \( \{x\} \) is the fractional part of \( x \), and for small values of \( x \), \( \{x\} \approx x \). Also, \( \lfloor x \rfloor = 0 \) for \( 0 < x < 1 \). Therefore, the function becomes:
\[ f(x) = \frac{\tan^2(x)}{x^2}. \] As \( x \to 0 \), \( \tan(x) \approx x \), so:
\[ f(x) = \frac{x^2}{x^2} = 1. \] Hence, the right-hand limit as \( x \to 0^+ \) is 1.
Left-hand limit (as \( x \to 0^- \)):
For \( x < 0 \), the function is:
\[ f(x) = \sqrt{\{x\} \cot\{x\}}. \] For negative \( x \), \( \{x\} = 1 - |x| \). As \( x \to 0^- \), \( \{x\} \to 1 \), and \( \cot(x) \to \infty \), making the expression under the square root diverge.
Thus, the left-hand limit does not exist.
Step 2: Conclusion
Since the left-hand limit does not exist, the overall limit of \( f(x) \) as \( x \to 0 \) does not exist.
Thus, the correct answer is:

\(\lim\limits_{x \to 0} f(x) \text{ does not exist}\)