If = lim x 0 e x 3 - ( 1 - x 3 ) 1 / 3 + ( ( 1 - x 2 ) 1 / 2 - 1 ) sin x x sin 2 x , then the value of 6 is ____________. [2022]

Question

If $β = \lim_{x \to 0} \frac{e^{x^{3}} - {(1 - x^{3})}^{1 / 3} + ({(1 - x^{2})}^{1 / 2} - 1) \sin x}{x \sin^{2} x},$ then the value of $6 β$ is ____________. [2022]

Smart Achievers · Accepted Answer

(5)

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

Using expansion,

$β = \lim_{x \to 0} [\frac{(1 + x^{3} + \frac{x^{6}}{2!} + \dots) - (1 - \frac{x^{3}}{3} + \frac{1}{3} \cdot \frac{- 2}{3} \cdot \frac{1}{2} x^{6} + \dots) + (- \frac{1}{2} x^{2} + \frac{1}{2} \cdot \frac{- 1}{2} \cdot \frac{1}{2} x^{4} + \dots) (x - \frac{x^{3}}{3!} + \dots)}{x^{3}}]$

$β = \lim_{x \to 0} \frac{x^{3} (1 + \frac{1}{3} - \frac{1}{2})}{x^{3}}$ (Neglecting higher powers of $x$ )

So, $β = \frac{5}{6}$

$∴$ $6 β = 5$

Solution

Q.
If $β = \lim_{x \to 0} \frac{e^{x^{3}} - {(1 - x^{3})}^{1 / 3} + ({(1 - x^{2})}^{1 / 2} - 1) \sin x}{x \sin^{2} x},$ then the value of $6 β$ is ____________. [2022]

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Solution

Q. If β=limx→0ex3-(1-x3)1/3+((1-x2)1/2-1)sinxxsin2x, then the value of 6β is ____________. [2022]

Ans. (5) β=limx→0ex3-(1-x3)1/3+((1-x2)1/2-1)sinxx·sin2xx2·x2 Using expansion, β=limx→0[(1+x3+x62!+⋯)-(1-x33+13·-23·12x6+⋯)+(-12x2+12·-12·12x4+⋯)(x-x33!+⋯)x3] β=limx→0x3(1+13-12)x3 (Neglecting higher powers of x) So, β=56 ∴ 6β=5

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Q.
If $β = \lim_{x \to 0} \frac{e^{x^{3}} - {(1 - x^{3})}^{1 / 3} + ({(1 - x^{2})}^{1 / 2} - 1) \sin x}{x \sin^{2} x},$ then the value of $6 β$ is ____________. [2022]