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

To evaluate the limit \( \lim_{x \to 0} \frac{\sin(2x) - 2\sin x}{x^3} \), we'll use known trigonometric identities and the concept of a Taylor series expansion for simplification.

1. Using the double angle identity, we have \( \sin(2x) = 2\sin x \cos x \).
2. Substitute \( \sin(2x) \) into the limit:

\[\lim_{x \to 0} \frac{2\sin x \cos x - 2\sin x}{x^3}\]

1. Factor out \( 2\sin x \) from the numerator:

\[\lim_{x \to 0} \frac{2\sin x (\cos x - 1)}{x^3}\]

1. For small \( x \), use the Taylor series approximations:
   • \( \sin x \approx x \)
   • \( \cos x \approx 1 - \frac{x^2}{2} \)
2. Substitute these approximations back into the limit:

\[\lim_{x \to 0} \frac{2x \left(-\frac{x^2}{2}\right)}{x^3}\]

1. Simplify the expression:

\(\lim_{x \to 0} \frac{-x^3}{x^3}  = -1\)

Therefore, the limit evaluates to \(-1\). Hence, the correct answer is: \( -1 \).