Question

find the above limit​

Answer 1

Answer:

a

Hope this helps.  Pls mark it Brainliest!!!! :D

Step-by-step explanation:

$\displaystyle \lim_{x\rightarrow 0} \frac{\sin ax}{x \cos bx}\\ \\= \lim_{x\rightarrow 0} \frac{\sin ax}{x}\frac{1}{\cos bx}\\ \\= a\left(\lim_{x\rightarrow 0} \frac{\sin ax}{ax}\right)\left(\lim_{x\rightarrow 0}\frac{1}{\cos bx}\right) \\ \\= a \times 1 \times 1 \\ \\= a$