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.