Find the limit of (sin 3x − 3 sin x)/x³ as x approaches 0.

Step-by-step solution

1. 1

   Confirm the indeterminate form and decide how deep to expand. At $x=0$ the numerator is $\sin 0-3\sin 0=0$ and the denominator is $0$, so this is $0/0$. Because the denominator is $x^{3}$, the numerator must be expanded to order $x^{3}$ — stopping at first order would only show that the leading terms cancel, without revealing the limit.

2. 2

   Expand $\sin 3x$ to third order. From $\sin u=u-\dfrac{u^{3}}{6}+O(u^{5})$ with $u=3x$,

   $\sin 3x=3x-\frac{27x^{3}}{6}+O(x^{5})=3x-\frac{9}{2}x^{3}+O(x^{5}).$

   The cube of the $3$ is what makes this term nine times larger than the corresponding term in $\sin x$ — that asymmetry is the whole source of the answer.

3. 3

   Expand $3\sin x$ to the same order.

   $3\sin x=3\left(x-\frac{x^{3}}{6}\right)+O(x^{5})=3x-\frac{1}{2}x^{3}+O(x^{5}).$

4. 4

   Subtract. The $3x$ terms cancel exactly, as the $0/0$ form promised:

   $\sin 3x-3\sin x=\left(-\frac92+\frac12\right)x^{3}+O(x^{5})=-4x^{3}+O(x^{5}).$

5. 5

   Divide and take the limit.

   $\frac{\sin 3x-3\sin x}{x^{3}}=-4+O(x^{2})\;\xrightarrow[x\to 0]{}\;-4.$

   Evaluating the quotient directly at $x=10^{-3}$ gives $-3.999998$, and at $x=10^{-4}$ gives $-3.99999998$, closing in on $-4$ from above at the predicted rate $O(x^{2})$.