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

Asked by Pragya Singh | 1 year ago |  377

##### 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 | 1 year ago

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