Question

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

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

Hint:

Always analyze the boundaries of \([x]\) and \(\{x\}\) individually based on whether you approach from the positive or negative direction before applying standard algebraic simplifications. A common mistake is assuming \(\{x\} \to 0\) from both sides, which leads to a false conclusion.

Answer 1

The Correct Option is D

Concept: This problem evaluates calculus limits involving special function properties:
• \([x]\) represents the Greatest Integer Function (floor function).
• \(\{x\}\) represents the Fractional Part Function, defined explicitly by the identity: \[ \{x\} = x - [x] \]
• For a limit to exist at a point \(x = a\), the Left-Hand Limit (\(\text{LHL}\)) must equal the Right-Hand Limit (\(\text{RHL}\)).
• Standard trigonometric limit theorem used: \(\lim_{\theta \to 0} \frac{\tan \theta}{\theta} = 1\).

Step 1: Evaluating the Right-Hand Limit ($\text{RHL}$) at $x \to 0^+$.
For the right-hand limit, \(x \to 0^+\), which means \(x\) is a tiny positive value (\(0 < x < 1\)). Under this condition: \[ [x] = 0 \quad \text{and} \quad \{x\} = x - [x] = x - 0 = x \] Substitute these expressions into the matching top branch of the piecewise definition: \[ \text{RHL} = \lim_{x \to 0^+} \frac{\tan^2\{x\}}{x^2 - [x]^2} = \lim_{x \to 0^+} \frac{\tan^2 x}{x^2 - 0^2} \] \[ \text{RHL} = \lim_{x \to 0^+} \left(\frac{\tan x}{x}\right)^2 = (1)^2 = 1 \]

Step 2: Evaluating the Left-Hand Limit ($\text{LHL}$) at $x \to 0^-$.
For the left-hand limit, \(x \to 0^-\), which means \(x\) is a tiny negative value (\(-1 < x < 0\)). Under this condition: \[ [x] = -1 \quad \text{and} \quad \{x\} = x - [x] = x - (-1) = x + 1 \] Substitute these expressions into the matching bottom branch of the piecewise definition: \[ \text{LHL} = \lim_{x \to 0^-} \sqrt{\{x\} \cot\{x\}} = \lim_{x \to 0^-} \sqrt{(x + 1) \cot(x + 1)} \] Since \(x \to 0\), substitute \(x = 0\) directly into this smooth continuous evaluation expression: \[ \text{LHL} = \sqrt{(0 + 1) \cot(0 + 1)} = \sqrt{1 \cdot \cot 1} = \sqrt{\cot 1} \]

Step 3: Comparing $\text{LHL}$ and $\text{RHL}$.
Comparing our calculated values: \[ \text{RHL} = 1 \quad \text{and} \quad \text{LHL} = \sqrt{\cot 1} \] Since \(\text{LHL} \neq \text{RHL}\), the two unidirectional pathways fail to meet at a single shared value, which means the overall limit does not exist: \[ \lim_{x \to 0} f(x) \quad \text{does not exist} \]