Question

check all that apply: $\frac{sin(2x)}{sin(9x)} =

$$square sinleft(\frac{2x}{9x} ight)$$

\boxed{checkmark} \frac{sin(2x)}{9x} \frac{9x}{sin(9x)}$$\boxed{checkmark} \frac{sin(2x)}{2x} \frac{2x}{sin(9x)}$$square \frac{2}{9}$$lim_{x \to 0} \frac{sin(2x)}{2x} = \boxed{1}$$lim_{x \to 0} \frac{2x}{sin(9x)} = \boxed{\frac{2}{9}}$$lim_{x \to 0} \frac{sin(2x)}{sin(9x)} = square$question help:submit questio

Explanation:

Step1: Rewrite the original expression

$$\frac{\sin(2x)}{\sin(9x)} = \frac{\sin(2x)}{2x} \cdot \frac{2x}{\sin(9x)}$$

Step2: Apply limit product rule

$$\lim_{x \to 0} \frac{\sin(2x)}{\sin(9x)} = \lim_{x \to 0} \frac{\sin(2x)}{2x} \cdot \lim_{x \to 0} \frac{2x}{\sin(9x)}$$

Step3: Substitute known limits

$$= 1 \cdot \frac{2}{9}$$

Answer:

$\frac{2}{9}$