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