Evaluate the given limit:  \(\lim\limits_{x \to 0} \dfrac{sinax+bx}{ax+sinbx} a,b,a+b\neq 0\)

Asked by Pragya Singh | 11 months ago |  356

1 Answer

Solution :-

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 | 11 months ago

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