Question

The value of
\[\lim_{n \to \infty} n \left( \sin \frac{1}{2n} - \frac{1}{2} e^{-\frac{1}{n}} + \frac{1}{2} \right) \]
is equal to __________ (answer in integer).

Answer 1

Correct Answer: 1

1. Expand \( \sin\frac{1}{2n} \): - Using the Taylor series expansion of \( \sin x \) around 0: \[ \sin\frac{1}{2n} \approx \frac{1}{2n} - \frac{1}{6}\left(\frac{1}{2n}\right)^3. \] 2. Expand \( e^{-\frac{1}{n}} \): - Using the Taylor series expansion of \( e^x \) around 0: \[ e^{-\frac{1}{n}} \approx 1 - \frac{1}{n} + \frac{1}{2n^2}. \] - Thus: \[ \frac{1}{2}e^{-\frac{1}{n}} \approx \frac{1}{2} - \frac{1}{2n} + \frac{1}{4n^2}. \] 3. Combine Terms: - Substitute the expansions into the expression: \[ n \left( \sin\frac{1}{2n} - \frac{1}{2}e^{-\frac{1}{n}} + \frac{1}{2} \right). \] - Simplify each term: \[ \sin\frac{1}{2n} \approx \frac{1}{2n}, \quad \frac{1}{2}e^{-\frac{1}{n}} \approx \frac{1}{2} - \frac{1}{2n}. \] - The expression becomes: \[ n \left( \frac{1}{2n} - \left(\frac{1}{2} - \frac{1}{2n}\right) + \frac{1}{2} \right). \] - Simplify: \[ n \left( \frac{1}{n} \right) = 1. \]