Question

Let \[ \alpha = \lim_{n \to \infty} \left(1 + n \sin \frac{3}{n^2}\right)^{2n}. \] Then, \(\ln \alpha\) is equal to ................

Hint:

For limits of the form \((1 + a/n)^{bn}\), the result tends to \(e^{ab}\), and \(\ln\) of the limit equals \(ab\).

Answer 1

Correct Answer: 6

Step 1: Simplify the expression inside the limit.
For small \(\theta\), \(\sin \theta \approx \theta\). Hence, \[ n \sin \frac{3}{n^2} \approx n \times \frac{3}{n^2} = \frac{3}{n}. \]
Step 2: Substitute into expression.
\[ \alpha = \lim_{n \to \infty} \left(1 + \frac{3}{n}\right)^{2n}. \]
Step 3: Take logarithm.
\[ \ln \alpha = \lim_{n \to \infty} 2n \ln\left(1 + \frac{3}{n}\right). \] Using \(\ln(1 + x) \approx x - \frac{x^2}{2}\) for small \(x\), \[ \ln \alpha = 2n \left(\frac{3}{n} - \frac{9}{2n^2}\right) = 6 - \frac{9}{n} \to 6. \] Final Answer: \[ \boxed{6} \]