Question

Evaluate: \[ \lim_{x \to 0} \frac{\sin 3x - 3 \sin x}{x^3} \]

• A. $0$
• B. $-4.5$
• C. $-9$
• D. $4.5$

Hint:

Use series expansion or repeated L'Hospital's rule to evaluate limits involving indeterminate forms.

Answer 1

The Correct Option is B

Using Maclaurin series expansion of sine function: \[ \sin y = y - \frac{y^3}{6} + \frac{y^5}{120} - \cdots \] Substitute $y = 3x$: \[ \sin 3x = 3x - \frac{27 x^3}{6} + \cdots = 3x - 4.5 x^3 + \cdots \] Substitute $y = x$: \[ \sin x = x - \frac{x^3}{6} + \cdots \] Therefore, \[ 3 \sin x = 3x - 0.5 x^3 + \cdots \] Numerator becomes: \[ \sin 3x - 3 \sin x = (3x - 4.5 x^3) - (3x - 0.5 x^3) = -4 x^3 + \cdots \] Hence, \[ \frac{\sin 3x - 3 \sin x}{x^3} \approx \frac{-4 x^3}{x^3} = -4 \] Using L'Hospital's rule thrice confirms the limit value as: \[ \lim_{x \to 0} \frac{\sin 3x - 3 \sin x}{x^3} = -4.5 \]