Question

Limx tends to 0 ax+xcosx/bsinx

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Answer 1

$\large\underline\blue{\bold{Given \:  Question :-  }}$

$\bf \:\lim_{x\to0} \dfrac{ax + xcosx}{bsinx}$

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$\bf \:\large \red{AηsωeR } ✍$

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$\large\underline\blue{\bold{Formula \: Used :-  }}$

$\bf \:\lim_{x\to0}\dfrac{sinx}{x} = 1$

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$\large\underline\purple{\bold{Solution :- }}$

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$\bf \:\lim_{x\to0} \dfrac{ax + xcosx}{bsinx}$

$\bf \: = \lim_{x\to0} \dfrac{x(a + cosx)}{bsinx}$

$\bf \: = \lim_{x\to0} \dfrac{a + cosx}{b \times \dfrac{sinx}{x} }$

$\bf \: =\dfrac{1}{b} \lim_{x\to0}(a + cosx) \times \bf \:\lim_{x\to0}\dfrac{1}{\dfrac{sinx}{x} }$

$\bf \: = \dfrac{1}{b} \times (a + 1) \times 1$

$\bf \: = \dfrac{a + 1}{b}$

$\large{\boxed{\boxed{\bf{Hence, \bf \:\lim_{x\to0} \dfrac{ax + xcosx}{bsinx} = \dfrac{a + 1}{b} }}}}$

Answer 2

Step-by-step explanation:

Iim x tends to 0 as+xcosx/bsinx = a+1/b