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To solve the problem, let’s analyze the given expression:

\[
\lim_{n \to \infty} \sum_{k=1}^{n} \frac{n - k}{n^2} \cos\left(\frac{4k}{n}\right)
\]

This expression resembles a Riemann sum. To convert it into an integral, we need to interpret the terms in the summation accordingly.

### Step-by-Step Solution:

1. **Rewrite the Sum as a Riemann Sum:**
   - As \( n \to \infty \), the sum can be interpreted as:
     \[
     \sum_{k=1}^{n} \frac{n - k}{n^2} \cos\left(\frac{4k}{n}\right)
     \approx \int_{0}^{1} (1 - x) \cos(4x) \, dx
     \]
   - Here, we let \( x = \frac{k}{n} \) so that \( \Delta x = \frac{1}{n} \).
   - Then, \( \frac{n - k}{n} = 1 - x \) and \( \frac{4k}{n} = 4x \).

2. **Set Up the Integral:**
   \[
   \int_{0}^{1} (1 - x) \cos(4x) \, dx
   \]

3. **Integrate by Parts:**
   To solve this integral, use integration by parts where:
   - Let \( u = 1 - x \) and \( dv = \cos(4x) \, dx \).
   - Then \( du = -dx \) and \( v = \frac{\sin(4x)}{4} \).

   Using the integration by parts formula \( \int u \, dv = uv - \int v \, du \):
   \[
   \int_{0}^{1} (1 - x) \cos(4x) \, dx = \left[ (1 - x) \frac{\sin(4x)}{4} \right]_{0}^{1} + \int_{0}^{1} \frac{\sin(4x)}{4} \, dx
   \]

4. **Evaluate Each Part:**
   - For the first term:
     \[
     \left[ (1 - x) \frac{\sin(4x)}{4} \right]_{0}^{1} = (0) - \left(\frac{1 \cdot \sin(0)}{4}\right) = 0
     \]
   - For the second term, integrate \( \frac{\sin(4x)}{4} \) from \( 0 \) to \( 1 \):
     \[
     \int_{0}^{1} \frac{\sin(4x)}{4} \, dx = \frac{1}{4} \int_{0}^{1} \sin(4x) \, dx = \frac{1}{4} \left[ -\frac{\cos(4x)}{4} \right]_{0}^{1}
     \]
     \[
     = \frac{1}{4} \cdot \left( -\frac{\cos(4)}{4} + \frac{1}{4} \right) = \frac{1}{16} (1 - \cos(4))
     \]

### Final Answer:
Thus, the value of the limit is:

\[
\frac{1}{16} (1 - \cos(4))
\]

### Correct Option:
The answer is **(d) \( \frac{1}{16} (1 - \cos 4) \)**.

Would you like further explanation or have any questions?

### Related Questions:
1. How does converting a summation to a Riemann integral help in finding limits?
2. Can integration by parts always be used with trigonometric functions?
3. What are some typical forms that indicate a problem can be solved using a Riemann sum?
4. How do we know which substitution to make when turning a discrete sum into an integral?
5. What is the significance of the \(\cos(4)\) term in this particular limit?

### Tip:
When working with Riemann sums, identifying the correct form of \( f(x) \) and \( dx \) in the context of the integral is key to solving the problem correctly.

The value of lim as n approaches infinity of the sum from k=1 to n of ((n-k)/n^2) * cos(4k/n) equals:

This problem explores the evaluation of a limit using Riemann sum approximation, involving the expression lim as n approaches infinity of the summation from k=1 to n of ((n-k)/n^2) * cos(4k/n). Learn how to convert the summation into an integral and solve using integration by parts. Suitable for undergraduate calculus students.

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