Math Problem Statement

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

Solution

The problem asks to compute the following limit:

limnn×((1+2n)n(1+1n)2n).\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:

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

Similarly:

limn(1+2n)n=e2.\lim_{n \to \infty} \left(1 + \frac{2}{n}\right)^n = e^2.

Thus, as nn \to \infty, we can expect the two expressions inside the parentheses to behave like e2e^2 and e2e^2, respectively.

Step 2: Use logarithmic approximations

For large nn, we can apply logarithmic approximations to the terms:

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

Thus:

  • (1+2n)ne2\left(1 + \frac{2}{n}\right)^n \approx e^2,
  • (1+1n)2ne2\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 nn is:

(1+2n)n(1+1n)2ne2e2=0.\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 nn becomes large, especially since we're multiplying by nn. 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 (1+2n)n\left(1 + \frac{2}{n}\right)^n:

ln(1+2n)2n222n2+O(1n3),\ln\left(1 + \frac{2}{n}\right) \approx \frac{2}{n} - \frac{2^2}{2n^2} + O\left(\frac{1}{n^3}\right),

so

(1+2n)ne2(12n+O(1n2)).\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 (1+1n)2n\left(1 + \frac{1}{n}\right)^{2n}:

ln(1+1n)1n12n2+O(1n3),\ln\left(1 + \frac{1}{n}\right) \approx \frac{1}{n} - \frac{1}{2n^2} + O\left(\frac{1}{n^3}\right),

so

(1+1n)2ne2(11n+O(1n2)).\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:

(1+2n)n(1+1n)2ne2(1n+O(1n2)).\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 nn

Multiplying the difference by nn, we get:

n×((1+2n)n(1+1n)2n)n×e2(1n)=e2.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:

limnn×((1+2n)n(1+1n)2n)=e2.\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.

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Math Problem Analysis

Mathematical Concepts

Calculus
Limits
Exponential Functions
Logarithmic Approximations

Formulas

Limit of \( \left(1 + \frac{1}{n}\right)^n \) as n approaches infinity is e
Asymptotic expansion using logarithmic approximations

Theorems

Standard limit theorem for exponential functions
Asymptotic behavior using series expansions

Suitable Grade Level

Undergraduate Level (Calculus II or III)