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Question #41883

Difficulty: 8/10 | 🟢 Free

The Question

Evaluate the limit: lim⁡x→0∑k=18∣tan⁡(kx)−sin⁡(kx)x2sin⁡(x)−108∣\lim_{x \to 0} \sum_{k=1}^{8} \left| \frac{\tan(kx) - \sin(kx)}{x^2 \sin(x)} - 108 \right|

Options

  • A.

    639

  • B.

    216

  • C.

    423

  • D.

    864

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

Trigonometric Standard Limits|Sigma Notation Properties|Modulus Sign Dependency

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to correctly handle the absolute value function in conjunction with limits and summations.

The primary trap lies in the temptation to commute the summation and the modulus, or to evaluate the sum algebraically before applying the absolute value, which yields the incorrect result of ∣−216∣=216|-216| = 216.

Another trap is using a first-order Taylor expansion (tan⁡(kx)≈kx\tan(kx) \approx kx, sin⁡(kx)≈kx\sin(kx) \approx kx), which incorrectly evaluates the limit to 0 and leads to the distractor 864.

Because the modulus creates a sign-flipping dependency based on the index kk, students must first evaluate the standard trigonometric limit for each term, identify the critical transition point (k=6k=6), and split the Sigma sum into 'positive' and 'negative' regions before computing the final arithmetic sum.

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterLimits and Derivatives - Class 11
ConceptTrigonometric Standard Limits
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree