Evaluate the given limit: $$\lim\limits_{x \to 0} \dfrac{ax+xcos\;x}{bsin\;x}$$

Asked by Pragya Singh | 11 months ago |  93

##### Solution :-

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

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 | 11 months ago

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