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Question #40352

Difficulty: 8/10 | 🟢 Free

The Question

Let $f: \mathbb{R} \to \mathbb{R}$ be a continuous function satisfying the inequality $2x^3 - 3x^4 \le f(x) \le 2x^3 + 3x^4$ for all $x \in (-1, 1)$ . A parametric curve is defined by $x(t) = \tan t - t$ and $y(t) = f(t)$ for $t \in (-1, 1)$ . Evaluate the limit: $\lim_{t \to 0} \frac{y(t)}{x(t)}$

Options

A.
6
B.
2
C.
0
D.
\text{Does not exist}

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Bridge

Practice problems that bridge the gap between foundational theory and advanced practical application.

Concepts Behind This Question

Polynomial Function Parity|Squeeze Theorem (Sandwich Theorem)|Parametric Derivative Endpoint Trap

Why This Question Is Tricky

Common Mistakes

This question tests the Squeeze Theorem while setting a lethal trap for students who over-rely on L'Hôpital's Rule for parametric limits.

The bounding polynomials $2t^3 \pm 3t^4$ restrict the function's value but not its derivative.

A valid function satisfying these bounds is $f(t) = 2t^3 + 3t^4 \sin(1/t^4)$ .

For this function, $f'(0) = 0$ , but for $t \neq 0$ , $f'(t)$ contains a term $-\frac{12}{t} \cos(1/t^4)$ , meaning $\lim_{t \to 0} f'(t)$ does not exist (the derivative is discontinuous at the origin).

Consequently, $\lim_{t \to 0} \frac{y'(t)}{x'(t)}$ does not exist.

Students who erroneously assume that the existence of a limit implies the validity of L'Hôpital's Rule will incorrectly choose 'Does not exist', failing to realize that the rule is inconclusive here and the Squeeze Theorem must be used directly on the function values.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Relations and Functions - Class 11
Concept	Polynomial Function Parity
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free