\( \lim\limits_{x \to 0} \dfrac{ax+xcos\;x}{bsin\;x} \)

At x = 0, the value of the given function takes the form

\(\dfrac{1}{b} \lim\limits_{x \to 0} \dfrac{x(a+cosx)}{sinx} \)

\( \dfrac{1}{b} \lim\limits_{x \to 0} (\dfrac{x}{sinx} )\times \lim\limits_{x \to 0}(a+cosx)\)

\( \dfrac{1}{b} \lim\limits_{x \to 0} (\dfrac{1}{ \lim\limits_{x \to 0} (\dfrac{sinx}{x})} )\times \lim\limits_{x \to 0}(a+cosx)\)

\( \dfrac{1}{b} \times {x \to 0}(a+cos0)\)

= \( \dfrac{a+1}{b}\)

Answered by Abhisek | 1 year ago