Question

The value of \(lim x.sin \frac {2}{x}\) is__________.

• A. 0
• B. 1
• C. 1/2
• D. 2

Answer 1

The Correct Option is D

To evaluate the limit \(\lim_{x \to \infty} x \cdot \sin\left(\frac{2}{x}\right)\), we'll use limit properties and trigonometric approximations.

1. Rewrite the Limit:
Let \( t = \frac{1}{x} \). As \( x \to \infty \), \( t \to 0^+ \).
The limit becomes: \[ \lim_{t \to 0^+} \frac{\sin(2t)}{t} \]

2. Apply Standard Limit:
Recall that: \[ \lim_{\theta \to 0} \frac{\sin\theta}{\theta} = 1 \] We can rewrite our expression to match this form: \[ \lim_{t \to 0^+} 2 \cdot \frac{\sin(2t)}{2t} = 2 \cdot 1 = 2 \]

3. Verification Using Series Expansion:
For small angles, \(\sin \theta \approx \theta - \frac{\theta^3}{6} + \cdots\). Thus: \[ \sin\left(\frac{2}{x}\right) \approx \frac{2}{x} - \frac{(2/x)^3}{6} + \cdots \] Multiplying by \(x\): \[ x \cdot \sin\left(\frac{2}{x}\right) \approx 2 - \frac{4}{3x^2} + \cdots \to 2 \text{ as } x \to \infty \]

4. Alternative Substitution:
Let \( y = \frac{2}{x} \), then as \( x \to \infty \), \( y \to 0 \): \[ \lim_{y \to 0} \frac{2}{y} \cdot \sin y = 2 \lim_{y \to 0} \frac{\sin y}{y} = 2 \times 1 = 2 \]

Final Answer:
The value of the limit is \(\boxed{2}\).