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The problem asks to compute the following limit:

\[
\lim_{n \to \infty} n \times \left( \left(1 + \frac{2}{n}\right)^n - \left(1 + \frac{1}{n}\right)^{2n} \right).
\]

### Step 1: Approximation using known limits
We know from standard limits that:

\[
\lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n = e.
\]

Similarly:

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

Thus, as \(n \to \infty\), we can expect the two expressions inside the parentheses to behave like \(e^2\) and \(e^2\), respectively.

### Step 2: Use logarithmic approximations
For large \(n\), we can apply logarithmic approximations to the terms:

\[
\left(1 + \frac{a}{n}\right)^n \approx e^a \quad \text{for large } n.
\]

Thus:

- \(\left(1 + \frac{2}{n}\right)^n \approx e^2\),
- \(\left(1 + \frac{1}{n}\right)^{2n} \approx e^2\).

### Step 3: Difference of the two terms
Therefore, the difference between the two terms for large \(n\) is:

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

However, we need to find the exact behavior of the difference as \(n\) becomes large, especially since we're multiplying by \(n\). This suggests a more careful asymptotic expansion of the terms.

### Step 4: Expansion using logarithms
We expand both terms using the first few terms of their logarithmic expansions.

For \(\left(1 + \frac{2}{n}\right)^n\):

\[
\ln\left(1 + \frac{2}{n}\right) \approx \frac{2}{n} - \frac{2^2}{2n^2} + O\left(\frac{1}{n^3}\right),
\]

so

\[
\left(1 + \frac{2}{n}\right)^n \approx e^2 \left(1 - \frac{2}{n} + O\left(\frac{1}{n^2}\right)\right).
\]

Similarly, for \(\left(1 + \frac{1}{n}\right)^{2n}\):

\[
\ln\left(1 + \frac{1}{n}\right) \approx \frac{1}{n} - \frac{1}{2n^2} + O\left(\frac{1}{n^3}\right),
\]

so

\[
\left(1 + \frac{1}{n}\right)^{2n} \approx e^2 \left(1 - \frac{1}{n} + O\left(\frac{1}{n^2}\right)\right).
\]

### Step 5: Finding the difference
Now we compute the difference:

\[
\left(1 + \frac{2}{n}\right)^n - \left(1 + \frac{1}{n}\right)^{2n} \approx e^2 \left(-\frac{1}{n} + O\left(\frac{1}{n^2}\right)\right).
\]

### Step 6: Multiplying by \(n\)
Multiplying the difference by \(n\), we get:

\[
n \times \left( \left(1 + \frac{2}{n}\right)^n - \left(1 + \frac{1}{n}\right)^{2n} \right) \approx n \times e^2 \left(-\frac{1}{n}\right) = -e^2.
\]

### Conclusion
Thus, the limit is:

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

Would you like further clarification or a deeper dive into any part of this solution?

Here are five related questions you might explore:
1. How do logarithmic expansions help in asymptotic analysis?
2. What is the role of the exponential function in limits involving powers of binomials?
3. Can you compute a similar limit using higher-order terms in the expansions?
4. How does L'Hopital's rule apply to limits like these?
5. What are the practical applications of such limits in real-world contexts?

**Tip**: When working with limits of exponential or binomial expressions, logarithmic approximations are a powerful tool for finding asymptotic behaviors.

Compute \( \lim_{n \to \infty} n \times \left( \left(1 + \frac{2}{n}\right)^n - \left(1 + \frac{1}{n}\right)^{2n} \right) \).

Learn how to compute the limit of \( n \times \left( \left(1 + \frac{2}{n}\right)^n - \left(1 + \frac{1}{n}\right)^{2n} \right) \) as n approaches infinity. This solution involves key concepts in calculus, including exponential functions, limits, and logarithmic approximations. Suitable for undergraduate students in Calculus II or III.

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