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To solve the given limit:

\[
\lim_{{n \to \infty}} \frac{-2(n!)}{n^n},
\]

we will analyze the behavior of the factorial \( n! \) and the term \( n^n \) as \( n \to \infty \).

### Step 1: Understanding \( n! \) and \( n^n \)
- \( n! \) (n factorial) is the product of all positive integers up to \( n \), which grows very rapidly.
- \( n^n \) represents \( n \) raised to the power of \( n \), which also grows quickly but at a different rate.

### Step 2: Use of Stirling's Approximation
Stirling's approximation can help estimate \( n! \), and it states:

\[
n! \approx \sqrt{2\pi n} \left(\frac{n}{e}\right)^n.
\]

Using this, we can rewrite the original limit as:

\[
\lim_{{n \to \infty}} \frac{-2 \sqrt{2\pi n} \left(\frac{n}{e}\right)^n}{n^n}.
\]

### Step 3: Simplifying the Expression
We can simplify the fraction:

\[
\frac{-2 \sqrt{2\pi n} \left(\frac{n}{e}\right)^n}{n^n} = -2 \sqrt{2\pi n} \left(\frac{n}{n e}\right)^n = -2 \sqrt{2\pi n} \left(\frac{1}{e}\right)^n.
\]

### Step 4: Evaluating the Limit
As \( n \to \infty \), \( \left(\frac{1}{e}\right)^n \to 0 \), because \( \frac{1}{e} < 1 \). Thus, the whole expression approaches zero:

\[
\lim_{{n \to \infty}} -2 \sqrt{2\pi n} \left(\frac{1}{e}\right)^n = 0.
\]

Therefore, the final result is:

\[
\lim_{{n \to \infty}} \frac{-2(n!)}{n^n} = 0.
\]

Would you like more details or have any questions?

Here are some related questions you could explore:
1. How does Stirling's approximation help in evaluating factorial limits?
2. What is the growth rate comparison between \( n! \) and \( n^n \)?
3. Can Stirling's approximation be used for other approximations involving factorials?
4. How does \( e \), the base of natural logarithms, play a role in such limits?
5. What would happen if the original expression had a different power in the denominator?

**Tip:** When dealing with factorials in limits, approximations like Stirling's can significantly simplify the problem.

Evaluate the limit: \( \lim_{{n \to \infty}} \frac{-2(n!)}{(n)^n} \)

Explore how to compute the limit of \( \lim_{{n \to \infty}} \frac{-2(n!)}{(n)^n} \), utilizing Stirling's approximation to simplify the factorial and exponential terms. This problem illustrates the difference in growth rates between factorials and powers, and uses advanced calculus techniques. Suitable for Calculus II students.

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