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Question #40480

Difficulty: 8/10 | 🟢 Free

The Question

Let z(x)=x2cos⁡(1x)+i3xz(x) = x^2 \cos\left(\frac{1}{x}\right) + i \sqrt{3} x for x∈R∖{0}x \in \mathbb{R} \setminus \{0\}. If f(x)=∣z(x)−1∣+∣z(x)+1∣f(x) = |z(x) - 1| + |z(x) + 1|, evaluate the limit: lim⁡x→0f(x)−2x2\lim_{x \to 0} \frac{f(x) - 2}{x^2}

Options

  • A.

    1

  • B.

    3\sqrt{3}

  • C.

    3

  • D.

    \text{Limit 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

Triangle Inequality Mind Games|Squeeze Theorem (Sandwich Theorem)|Non-Differentiable Extrema

Why This Question Is Tricky

Common Mistakes

This question sets a lethal trap for students overly reliant on L'Hopital's Rule.

The limit evaluates to a 0/00/0 indeterminate form.

If a student attempts to differentiate the numerator, they must compute the derivative of u(x)=x2cos⁡(1/x)u(x) = x^2 \cos(1/x), which is 2xcos⁡(1/x)+sin⁡(1/x)2x \cos(1/x) + \sin(1/x).

As x→0x \to 0, the sin⁡(1/x)\sin(1/x) term oscillates wildly, meaning the limit of the derivative does not exist.

This baits the student into incorrectly choosing 'Limit does not exist'.

However, L'Hopital's Rule only states that if the limit of the derivatives exists, it equals the original limit; its failure does not imply the original limit diverges.

By leveraging the geometric properties of the complex plane and applying the Squeeze Theorem, the limit is rigorously sandwiched to exactly 3.

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterComplex Numbers and Quadratic Equations - Class 11
ConceptTriangle Inequality Mind Games
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree