Question

`\rm If \: \beta = \lim_(x \to0) \frac{ {e}^{ {x}^(3) } - (1 - {x}^(3) {)}^{ \frac{1}3 } + ((1 - {x}^(2) {)}^{ (1)/(2) } - 1) \sin x}{x \sin^(2) x} ,\\`

then the value of
`6 \beta` is​

Answer 1

Expand the limand as

`(e^(x^3) - (1 - x^3)^(1/3) + \left((1-x^2)^(1/2) - 1\right) \sin(x))/(x \sin^2(x)) \\\\ ~~~~~~~~ = (e^(x^3) - (1-x^3)^(1/3))/(x^3) \cdot (x^2)/(\sin^2(x)) + ((1-x^2)^(1/2) - 1)/(x^2) \cdot (x)/(\sin(x))`

Recall that for
`a\\eq0`,

`\displaystyle \lim_(x\to0) (\sin(ax))/(ax) = 1`

Then

`\displaystyle \beta = \lim_(x\to0) \left( (e^(x^3) - (1-x^3)^(1/3))/(x^3) + ((1-x^2)^(1/2) - 1)/(x^2)\right)`

Rationalize the numerators.

`(e^(x^3) - (1-x^3)^(1/3))/(x^3) = (e^(3x^3) - (1 - x^3))/(x^3 \left(e^(2x^3) + e^(x^3) (1-x^3)^(1/3) + (1-x^3)^(2/3)\right)) \\\\ ~~~~~~~~~~~ = \left((e^(3x^3) - 1)/(x^3) + 1\right) \cdot\frac1{e^(2x^3) + e^(x^3) (1-x^3)^(1/3) + (1-x^3)^(2/3)}`

and observe that

`\frac1{e^(2x^3) + e^(x^3) (1-x^3)^(1/3) + (1-x^3)^(2/3)} \to \frac13`

`((1-x^2)^(1/2)-1)/(x^2) = ((1-x^2) - 1)/(x^2 \left((1-x^2)^(1/2) + 1\right)) \\\\ ~~~~~~~~ = -\frac1{(1-x^2)^(1/2) + 1} \to -\frac12`

This leaves us with

`\displaystyle \beta = \lim_(x\to0) (e^(3x^3) - 1)/(3x^3) - \frac16`

The remaining limit is the derivative of
`e^(3x^3)` at
`x=0`, so its value is 1, and we find

`\beta = 1 - \frac16 = \frac56 \implies 6\beta = \boxed{5}`