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Question #41309

Difficulty: 8/10 | 🟢 Free

The Question

Let $f: \mathbb{R} \to \mathbb{R}$ be defined as $f(x) = \begin{cases} \lim_{n \to \infty} \frac{x^2(3+\sin(1/x))}{1 + x^{2n}}, & x \neq 0 \\ 0, & x = 0 \end{cases}$ Let $g(x) = \text{sgn}(f'(x))$ for $x \in (-1, 1) \setminus \{0\}$ . Evaluate the pair $(L_1, L_2)$ , where $L_1 = \limsup_{x \to 0^-} g(x) \quad \text{and} \quad L_2 = \liminf_{x \to 0^-} g(x)$

Options

A.
$(1, -1)$
B.
$(-1, -1)$
C.
$(0, 0)$
D.
$(1, 1)$

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Monster

Ultra-high difficulty questions designed to test your limit, stamina, and deep conceptual understanding.

Concepts Behind This Question

Composition Continuity|Squeeze Theorem (Sandwich Theorem)|First Derivative Test Trap

Why This Question Is Tricky

Common Mistakes

This question tests the interaction between the Squeeze Theorem and the First Derivative Test trap.

Students often correctly identify that $f(x) \ge 2x^2 > 0$ for $x \neq 0$ and $f(0)=0$ , making $x=0$ a strict local minimum.

The trap lies in assuming that a local minimum implies the function must be strictly decreasing on its immediate left (i.e., $f'(x) < 0$ as $x \to 0^-$ ).

This monotonicity trap leads them to incorrectly conclude that $\text{sgn}(f'(x)) = -1$ as $x \to 0^-$ , yielding the incorrect pair $(-1, -1)$ .

However, due to the oscillating $\cos(1/x)$ term dominating the derivative near zero, $f(x)$ is not monotonic in any neighborhood of $0$ , causing the signum composition to oscillate and the limits to diverge.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Continuity and Differentiability - Class 12
Concept	Composition Continuity
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free