Evaluating lim sin(3x)/x using the standard sine limit

Rewrite the expression by multiplying and dividing by 3: $\frac{\sin(3x)}{x} = 3 \cdot \frac{\sin(3x)}{3x}.$ This is valid for all $x \neq 0$, so the rewritten function equals the original one near $x=0$.

Let $\theta = 3x$; as $x \to 0$, $\theta \to 0$ as well, so $\dfrac{\sin(3x)}{3x} \to 1$ by the known limit. Applying the product rule for limits, $\lim_{x\to 0} 3\cdot\frac{\sin(3x)}{3x} = 3 \cdot \lim_{x\to 0}\frac{\sin(3x)}{3x} = 3 \cdot 1 = 3.$ Therefore the limit equals 3.

The answer is three. Since the argument of sine was three x, the expression sine of three x over x was rewritten as three times sine of three x over three x, which let theta equal three x match the standard limit form.