show that the limit as n→∞ of the function (nr^2sin(2π/n))/2cos(2π/n) is equal to πr^2 if we let k = 2π/n

We are tasked with showing that the limit of the expression:

\[
\lim_{n \to \infty} \frac{n r^2 \sin\left(\frac{2\pi}{n}\right)}{2 \cos\left(\frac{2\pi}{n}\right)}
\]

is equal to \( \pi r^2 \) by letting \( k = \frac{2\pi}{n} \).

### Step 1: Substituting \( k = \frac{2\pi}{n} \)
First, express the given expression in terms of \( k \).

Since \( k = \frac{2\pi}{n} \), this implies that \( n = \frac{2\pi}{k} \).

Now substitute this into the original expression:

\[
\frac{n r^2 \sin\left(\frac{2\pi}{n}\right)}{2 \cos\left(\frac{2\pi}{n}\right)} = \frac{\left(\frac{2\pi}{k}\right) r^2 \sin(k)}{2 \cos(k)}
\]

This simplifies to:

\[
\frac{r^2 \cdot 2\pi \sin(k)}{2k \cos(k)}
\]

which reduces to:

\[
\frac{\pi r^2 \sin(k)}{k \cos(k)}
\]

### Step 2: Taking the limit as \( n \to \infty \)
As \( n \to \infty \), \( k = \frac{2\pi}{n} \) implies that \( k \to 0 \). Thus, we now need to evaluate the limit:

\[
\lim_{k \to 0} \frac{\pi r^2 \sin(k)}{k \cos(k)}
\]

### Step 3: Evaluating the limit
To solve this limit, we can break it into two parts:

\[
\lim_{k \to 0} \frac{\sin(k)}{k} \quad \text{and} \quad \lim_{k \to 0} \cos(k)
\]

#### Part 1: \( \lim_{k \to 0} \frac{\sin(k)}{k} \)
This is a standard trigonometric limit, and it is well known that:

\[
\lim_{k \to 0} \frac{\sin(k)}{k} = 1
\]

#### Part 2: \( \lim_{k \to 0} \cos(k) \)
This is also a standard limit, and we know that:

\[
\lim_{k \to 0} \cos(k) = 1
\]

### Step 4: Applying the limits
Now, applying these limits to the expression:

\[
\lim_{k \to 0} \frac{\pi r^2 \sin(k)}{k \cos(k)} = \pi r^2 \cdot \frac{1}{1} = \pi r^2
\]

Thus, we have shown that:

\[
\lim_{n \to \infty} \frac{n r^2 \sin\left(\frac{2\pi}{n}\right)}{2 \cos\left(\frac{2\pi}{n}\right)} = \pi r^2
\]

### Conclusion
The limit is indeed \( \pi r^2 \).

---

Do you want further details on any step or have any questions?

Here are five questions for further exploration:
1. Why is \( \lim_{k \to 0} \frac{\sin(k)}{k} = 1 \)?
2. What is the geometric interpretation of this limit?
3. How does L'Hopital's Rule apply to this type of limit problem?
4. Can we apply a Taylor series expansion to analyze the limit?
5. How do small angle approximations simplify trigonometric limits?

**Tip:** When working with limits involving trigonometric functions, remembering basic limits like \( \lim_{x \to 0} \frac{\sin(x)}{x} = 1 \) can be very useful.

Show that the limit as n→∞ of the function (nr^2sin(2π/n))/2cos(2π/n) is equal to πr^2 if we let k = 2π/n.

This problem demonstrates how to compute the limit of (nr^2sin(2π/n))/(2cos(2π/n)) as n approaches infinity, using trigonometric limit theorems. By substituting k = 2π/n, the limit is simplified to πr^2. Suitable for university-level calculus students.

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