Question

The value of \[ \lim_{n \to \infty} 8n \left( \left( e^{\frac{1}{2n}} - 1 \right) \left( \sin \frac{1}{2n} + \cos \frac{1}{2n} \right) \right) \] is equal to ............... (rounded off to two decimal places).

Hint:

When evaluating limits involving small angles or terms, use standard approximations like \( e^x - 1 \approx x \) and \( \sin x \approx x \), \( \cos x \approx 1 \) for small \( x \).

Answer 1

Step 1: Simplify the expression for small \( n \).
For large \( n \), \( \frac{1}{n} \) becomes small, so we can use the approximations: \[ e^{x} - 1 \approx x \quad \text{for small } x, \] \[ \sin x \approx x \quad \text{and} \quad \cos x \approx 1 \quad \text{for small } x. \] Substituting these approximations into the given limit expression, we get: \[ e^{\frac{1}{2n}} - 1 \approx \frac{1}{2n}, \quad \sin \frac{1}{2n} \approx \frac{1}{2n}, \quad \cos \frac{1}{2n} \approx 1. \] Thus, the limit simplifies to: \[ \lim_{n \to \infty} 8n \left( \frac{1}{2n} \left( \frac{1}{2n} + 1 \right) \right) = \lim_{n \to \infty} 8n \cdot \frac{1}{2n} \cdot \left( \frac{1}{2n} + 1 \right). \] Step 2: Evaluate the limit.
Now simplify the expression: \[ 8n \cdot \frac{1}{2n} \cdot \left( \frac{1}{2n} + 1 \right) = 4 \left( \frac{1}{2n} + 1 \right) = 4 \left( 0 + 1 \right) = 4. \] Thus, the value of the limit is 4. Final Answer: \[ \boxed{4}. \]