If lim x - 0a sin x-b x-c x2-x3-2 x2log 1-x-2 x3-x4 exists and is finite- then

The correct option is C c = 0 x → 0lima(x x^3/6+x^5/120 .....) bx+cx^2+x^3/2x^2(x x^2/2+x^3/3 ......) 2x^3+x^4 =x → 0lim(a b)x+cx^2+(1 a/6)x^3+ax^5/120+..../2x ...

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Byju's Answer

Standard XII

Mathematics

Sum of n Terms

If lim x → 0a...

Question

If $lim x \to 0 \frac{a sin x - b x + c x^{2} + x^{3}}{2 x^{2} log (1 + x) - 2 x^{3} + x^{4}}$ 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
$a = 6$

$lim x \to 0 \frac{a (x - \frac{x^{3}}{6} + \frac{x^{5}}{120} - . . . . .) - b x + c x^{2} + x^{3}}{2 x^{2} (x - \frac{x^{2}}{2} + \frac{x^{3}}{3} - . . . . . .) - 2 x^{3} + x^{4}}$

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

For this limit to exist we must have $a - b = 0, c = 0,$ and $1 - \frac{a}{6} = 0$ , that is, $a = b = 6$ and $c = 0$

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If limx→0asinx−bx+cx2+x32x2log(1+x)−2x3+x4 exists and is finite, then

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