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?
Best Answer
A bit of better than your: $$\lim_{x\rightarrow0}\frac {\sin ax\cos bx}{\sin cx}=\lim_{x\rightarrow0}\left(\cos{bx}\cdot\frac{\sin{ax}}{ax}\cdot\frac{cx}{\sin{cx}}\cdot\frac{a}{c}\right)=\frac{a}{c}$$