Question

Evaluate the limit: \[ \lim_{x \to 0} \frac{\sin(2x) - 2\sin x}{x^3} \]

• A. \( 1 \)
• B. \( -1 \)
• C. \( 0 \)
• D. \( 2 \)

Hint:

For limits involving trigonometric functions as \( x \to 0 \), use Taylor expansions or standard limits to simplify higher-order terms.

Answer 1

The Correct Option is B

Step 1: Use the Taylor series expansions near \( x = 0 \): \[ \sin x = x - \frac{x^3}{6} + \cdots \] \[ \sin(2x) = 2x - \frac{(2x)^3}{6} + \cdots = 2x - \frac{8x^3}{6} + \cdots \]
Step 2: Substitute into the given expression: \[ \sin(2x) - 2\sin x = \left(2x - \frac{8x^3}{6}\right) - 2\left(x - \frac{x^3}{6}\right) \]
Step 3: Simplify: \[ = 2x - \frac{8x^3}{6} - 2x + \frac{2x^3}{6} = -\frac{6x^3}{6} = -x^3 \]
Step 4: Divide by \( x^3 \) and take the limit: \[ \lim_{x \to 0} \frac{-x^3}{x^3} = -1 \]