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Question #41912

Difficulty: 8/10 | 🟢 Free

The Question

Evaluate the following limit: $\lim_{n \to \infty} \left( \frac{1}{n^2} \sum_{k=1}^{n} k \cos(k) + \sum_{k=1}^{n} \cos\left(\frac{k}{n}\right) \sin\left(\frac{k}{n^2}\right) \right)$

Options

A.
$\sin 1 + \cos 1 - 1$
B.
$\sin 1 - \cos 1$
C.
$\cos 1 - \sin 1 + 1$
D.
\text{The limit 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

The L'Hôpital's Rule Misuse|Trigonometric Standard Limits|Sum to n terms vs. General Term

Why This Question Is Tricky

Common Mistakes

This question tests the 'Lethal Collision' between discrete summation limits and continuous calculus rules.

Students often misuse Stolz-Cesàro (the discrete L'Hôpital's rule) or convert the sum to an integral without verifying conditions.

For the first term, applying Stolz-Cesàro yields $\lim \frac{(n+1)\cos(n+1)}{2n+1} \approx \frac{1}{2}\cos(n+1)$ , leading to the false conclusion that the limit oscillates and does not exist.

However, summation by parts proves it converges to 0.

The second term requires recognizing that $\sin(k/n^2)$ can be linearized to $k/n^2$ because the error term vanishes, transforming it into a standard Riemann sum.

Failing to separate the bounded oscillatory growth from the Riemann sum traps students into choosing 'The limit does not exist'.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Continuity and Differentiability - Class 12
Concept	The L'Hôpital's Rule Misuse
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free