Question

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

• A. $-\frac{1}{6}$
• B. $\frac{1}{6}$
• C. $0$
• D. $\infty$

Hint:

For limits involving $\sin x$, consider L’Hôpital’s Rule or Taylor series expansion when encountering indeterminate forms.

Answer 1

The Correct Option is A

Substituting $x = 0$ gives:
\[ \frac{\sin 0 - 0}{0^3} = \frac{0}{0} \]
This is an indeterminate form, so we apply L’Hôpital’s Rule.
Differentiate numerator and denominator:
\[ \text{Numerator: } \frac{d}{dx} (\sin x - x) = \cos x - 1 \]
\[ \text{Denominator: } \frac{d}{dx} (x^3) = 3x^2 \]
\[ \lim_{x \to 0} \frac{\cos x - 1}{3x^2} \]
At $x = 0$, this is still $\frac{0}{0}$. Apply L’Hôpital’s Rule again:
\[ \text{Numerator: } \frac{d}{dx} (\cos x - 1) = -\sin x \]
\[ \text{Denominator: } \frac{d}{dx} (3x^2) = 6x \]
\[ \lim_{x \to 0} \frac{-\sin x}{6x} = \lim_{x \to 0} -\frac{\sin x}{x} \cdot \frac{1}{6} = -\frac{1}{6} \cdot 1 = -\frac{1}{6} \]
Alternatively, use the Taylor series for $\sin x$:
\[ \sin x = x - \frac{x^3}{6} + O(x^5) \]
\[ \frac{\sin x - x}{x^3} = \frac{(x - \frac{x^3}{6} - x)}{x^3} = \frac{-\frac{x^3}{6}}{x^3} = -\frac{1}{6} \]
Both methods confirm the limit is $-\frac{1}{6}$.