The value of a,b and c such that If limx→0ax ex−blog(1+x)+cxe−xx2sinx=2

limx→0cx(1−x1!+x22!+x33!)x2(x−x33!+x35!)Coefficient of x & x2=0 and coefficient of x3=2

The value of a,b and c such that If limx→0ax ex−blog(1+x)+cxe−xx2sinx=2

limx→0cx(1−x1!+x22!+x33!)x2(x−x33!+x35!)Coefficient of x & x2=0 and coefficient of x3=2

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The value of a,b and c such that $I f \lim_{x \to 0} \frac{a x e^{x} - b \log (1 + x) + c x e^{- x}}{x^{2} \sin x} = 2$

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a=3, b=12, c=9

a=3,b=-12,c=-9

a=1,b=2,c=4

a=2,b=10,c=8

answer is A.

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Detailed Solution

$\lim_{x \to 0} \frac{c x (1 - \frac{x}{1!} + \frac{x^{2}}{2!} + \frac{x^{3}}{3!})}{x^{2} (x - \frac{x^{3}}{3!} + \frac{x^{3}}{5!})}$

Coefficient of x & $x^{2} = 0$ and coefficient of $x^{3} = 2$

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The value of a,b and c such that If limx→0ax ex−blog(1+x)+cxe−xx2sinx=2

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The value of a,b and c such that $I f \lim_{x \to 0} \frac{a x e^{x} - b \log (1 + x) + c x e^{- x}}{x^{2} \sin x} = 2$