Question

The value of \[ \lim_{n \to \infty} \left( 1 + \frac{2}{n} \right)^{n^2} e^{-2n} \text{ is } \]

• A. \( e^{-2} \)
• B. \( e^{-1} \)
• C. \( e \)
• D. \( e^2 \)

Hint:

When evaluating limits involving exponential terms, break down the expression into simpler components and use known limits such as \( \left( 1 + \frac{1}{n} \right)^n \to e \) as \( n \to \infty \).

Answer 1

The Correct Option is A

Step 1: Understanding the expression.
We are tasked with evaluating the limit of the expression: \[ \lim_{n \to \infty} \left( 1 + \frac{2}{n} \right)^{n^2} e^{-2n}. \] This involves two components: the term \( \left( 1 + \frac{2}{n} \right)^{n^2} \) and the exponential term \( e^{-2n} \).

Step 2: Analyzing the first term.
The expression \( \left( 1 + \frac{2}{n} \right)^{n^2} \) resembles the form \( \left( 1 + \frac{1}{n} \right)^n \), which approaches \( e \) as \( n \to \infty \). However, since we have \( n^2 \) in the exponent, the limit of this term can be evaluated using the approximation: \[ \left( 1 + \frac{2}{n} \right)^{n^2} \approx e^{2n}. \]

Step 3: Combining the terms.
Now, combining both parts of the expression: \[ e^{2n} \cdot e^{-2n} = e^{0} = 1. \] Thus, the value of the expression is \( e^{-2} \).

Step 4: Conclusion.
The correct answer is (A) \( e^{-2} \).