Question

Evaluate the limit: \[ \lim_{n \to \infty} \frac{(2^n + n 2^n \sin^2 \frac{n}{2})}{(2n - n \cos \frac{1}{n})} \] The value of the limit is _________ (round off to 2 decimal places).

Hint:

For limits involving trigonometric functions with high powers of \( n \), consider the dominant terms and simplify the expression before evaluating the limit.

Answer 1

Correct Answer: 2

The expression inside the limit is complex, but we can analyze it by considering the behavior of each term as \( n \to \infty \). First, look at the asymptotic behavior of each part of the expression:
\[ \lim_{n \to \infty} \frac{(2^n + n 2^n \sin^2 \frac{n}{2})}{(2n - n \cos \frac{1}{n})} \] For large \( n \), \( \sin^2 \frac{n}{2} \) oscillates between 0 and 1, and \( \cos \frac{1}{n} \to 1 \). Hence, the dominant terms will be the powers of \( 2^n \). The expression simplifies to: \[ \frac{2^n (1 + n \sin^2 \frac{n}{2})}{2n}. \] Since the factor \( \sin^2 \frac{n}{2} \) is bounded, the overall limit approaches: \[ \frac{2^n}{2n} \quad \text{as } n \to \infty. \] Thus, the value of the limit approaches a large number, which can be computed more precisely.