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Question #41568

Difficulty: 8/10 | 🟢 Free

The Question

A parametric curve is defined by $x(t)$ and $y(t)$ for $t > 0$ , where: $x(t) = \lim_{n \to \infty} \frac{t}{n} \sum_{k=1}^n \left[ 2\left(\frac{kt}{n}\right) \sin\left(\frac{n}{kt}\right) - \cos\left(\frac{n}{kt}\right) + 1 \right]$ $y(t) = \lim_{n \to \infty} \frac{t}{n} \sum_{k=1}^n \left[ 2\left(\frac{kt}{n}\right) \sin\left(\frac{n}{kt}\right) - \cos\left(\frac{n}{kt}\right) + 7 \right]$ Evaluate $\lim_{t \to 0^+} \frac{y(t)}{x(t)}$ .

Options

A.
7
B.
1
C.
0
D.
\text{Does not exist}

Check Your Answer

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

Parametric Curve Area|Integral as the Limit of a Sum|Squeeze Theorem (Sandwich Theorem)|L'Hopital's Pitfall

Why This Question Is Tricky

Common Mistakes

The question tests the evaluation of limits involving integrals of oscillating functions.

The primary trap is the premature application of L'Hopital's rule.

If a student differentiates the integrals using Leibniz's rule, they obtain $\lim_{t \to 0^+} \frac{y'(t)}{x'(t)} = \lim_{t \to 0^+} \frac{2t \sin(1/t) - \cos(1/t) + 7}{2t \sin(1/t) - \cos(1/t) + 1}$ .

Because $x'(t)$ periodically equals $0$ as $t \to 0$ (e.g., at $t = \frac{1}{2m\pi}$ ), this leads to a 'division by zero' in the derivative ratio, violating L'Hopital's condition that $x'(t) \neq 0$ in a punctured neighborhood.

This tricks students into concluding the limit does not exist.

The correct path requires evaluating the integral first and applying the Squeeze Theorem.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Application of Integrals - Class 12
Concept	Parametric Curve Area
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free