The value of the limit limx→0ex3−(1−x3)1/3+((1−x2)1/2−1)sinxxsinx . tanx is equal to

β=limx→01+x3−1+13x3+((1−12x2−1))sinxxsinx.tanxβ=limx→043x3−12x2(x)x3=43−12=8−36=56

The value of the limit limx→0ex3−(1−x3)1/3+((1−x2)1/2−1)sinxxsinx . tanx is equal to

β=limx→01+x3−1+13x3+((1−12x2−1))sinxxsinx.tanxβ=limx→043x3−12x2(x)x3=43−12=8−36=56

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The value of the limit $\lim_{x \to 0 } \frac{e^{x^{3}} - {(1 - x^{3})}^{1 / 3} + ({(1 - x^{2})}^{1 / 2} - 1) \sin x}{x \sin x . \tan x}$ is equal to

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$\frac{3}{2}$

$\frac{2}{3}$

$\frac{5}{6}$

$\frac{4}{5}$

answer is C.

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Detailed Solution

$β = \lim_{x \to 0} \frac{1 + x^{3} - 1 + \frac{1}{3} x^{3} + ((1 - \frac{1}{2} x^{2} - 1)) \sin x}{x \sin x . \tan x}$

$β = \lim_{x \to 0} \frac{\frac{4}{3} x^{3} - \frac{1}{2} x^{2} (x)}{x^{3}} = \frac{4}{3} - \frac{1}{2} = \frac{8 - 3}{6} = \frac{5}{6}$

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