\( \lim\limits_{x \to 0} \dfrac{sinax}{sinbx},a,b\neq 0\)
At x = 0, the value of the given function takes the form \( \dfrac{0}{0}\)
= \( \lim\limits_{x \to 0} \dfrac{\dfrac{sinax}{ax}\times ax}{\dfrac{sin\;bx}{ax}\times bx}\)
= \( \dfrac{a}{b}\times \dfrac{ \lim\limits_{ax \to 1}\dfrac{sinax}{ax}}{ \lim\limits_{bx \to 1} \dfrac{sin\;bx}{ax}}\)
= \( \dfrac{a}{b}\times \dfrac{1}{1}\)
= \( \dfrac{a}{b}\)
Answered by Pragya Singh | 1 year ago