$lim x \to 0 \frac{a sin x - b x + c x^{2} + x^{3}}{2 x^{2} log (1 + x) - 2 x^{3} + x^{4}}$ exists and is finite, then

a = b

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c = 0

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a = 6

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c = 2

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Solution

The correct options are
A $a = b$
B $c = 0$
C $a = 6$
$lim x \to 0 \frac{a (x - \frac{x^{3}}{6} + \frac{x^{5}}{120} - . . . .) - b x + c x^{2} + x^{3}}{2 x^{2} (x - \frac{x^{2}}{2} + \frac{x^{3}}{3} - . . . . . . . . .) - 2 x^{3} + x^{4}}$

$= lim x \to 0 \frac{(a - b) x + c x^{2} + (1 - a / 6) x^{3} + a x^{5} / 120 + . . .}{2 x^{5} / 3 - x^{6} / 2 + . . .}$

For this limit to exist we must have $a - b = 0, c = 0$ , and $1 - a / 6 = 0$ , that is, $a = b = 6$ and $c = 0.$

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