Calculate $\lim_{x\to0}\frac{\sin ax\cos bx}{\sin cx}$ Calculate: $$\lim_{x\to0} \frac{\sin ax\cos bx}{\sin cx}$$


My attempt:
$$=\lim_{x\to 0} \dfrac {\sin (ax). \cos (bx)}{\sin (cx)}$$
$$=\lim_{x\to 0} \dfrac {\sin (ax)}{ax} \times \dfrac {ax}{\sin (cx)} \times cx \times \dfrac {\cos (bx)}{cx}$$
$$=1\times ax \times 1 \times \dfrac {\cos (bx)}{cx}$$

How do I do further?
A bit of better than your:
$$\lim_{x\rightarrow0}\frac {\sin ax\cos bx}{\sin cx}=\lim_{x\rightarrow0}\left(\cos{bx}\cdot\frac{\sin{ax}}{ax}\cdot\frac{cx}{\sin{cx}}\cdot\frac{a}{c}\right)=\frac{a}{c}$$
$$\lim_{x \rightarrow 0}\frac{\sin \text ax \cos \text bx}{\sin \text cx}$$

Using linear approximation

$$L=f(\text a)+f'(x)(x-\text a)$$

Therefore, at $x=0$
$$\sin \text ax\approx \text ax$$
$$\sin \text cx\approx \text cx$$
$$\cos bx \approx 1$$

$$\lim_{x \rightarrow 0}\frac{\sin \text ax \cos \text bx}{\sin \text cx}=\lim_{x \rightarrow 0}\frac{(\text ax)(1)}{\text cx} \iff \lim_{x \rightarrow 0}\frac{(\text ax)(1)}{\text cx}=\lim_{x \rightarrow 0} \frac{\text a}{\text c}$$

$$\therefore \lim_{x \rightarrow 0} \frac{\text a}{\text c}=\frac{\text a}{\text c}$$