At x = 0, the value of the given function takes the form \( \dfrac{0}{0}\)
\( \lim\limits_{x \to 0} \dfrac{sinax+bx}{ax+sinbx} \)
\( \lim\limits_{x \to 0} \dfrac{(\dfrac{sinax}{ax})ax+bx}{ax+bx(\dfrac{sinbx}{bx})}\)
= \( \dfrac{ \lim\limits_{x \to 0}(ax)+\lim\limits_{x \to 0}bx)}{ \lim\limits_{x \to 0}(ax)+\lim\limits_{x \to 0}(bx)}\)
= \( \dfrac{ \lim\limits_{x \to 0}(ax+bx)}{\lim\limits_{x \to 0}(ax+bx)}\)
= \( \lim\limits_{x \to 0}(1)\)
= 1
Answered by Abhisek | 1 year ago