Question

If
$\beta =\underset{x\to 0}{\lim }\frac{e^{x^3}-{\left(1-x^3\right)}^{\frac{1}{3}}+\left({\left(1-x^2\right)}^{\frac{1}{2}}-1\right)\sin x}{x{\sin }^{2}x}$
then the value of $6\beta$ is _____

Answer 1

Answer: 5

$\beta =\underset{x\to 0}{\lim }\frac{e^{x^3}-{\left(1-x^3\right)}^{1/3}}{\frac{x{\sin }^{2}x}{x^2}{x}^2}+\frac{\left({\left(1-x^2\right)}^{1/2}-1\right)\sin x}{x\frac{{\sin }^{2}x}{x^2}{x}^2}$
use expansion
$\beta =\underset{x\to 0}{\lim }\frac{\left(1+x^3\right)-\left(1-\frac{x^3}{3}\right)}{x^3}+\underset{x\to 0}{\lim }\frac{\left(\left(1-\frac{x^2}{2}\right)-1\right)}{x^2}\frac{\sin x}{x}$
$\beta =\underset{x\to 0}{\lim }\frac{4x^3}{3x^3}+\underset{x\to 0}{\lim }\frac{-x^2}{2x^2}$
$\beta =\frac{4}{3}-\frac{1}{2}=\frac{5}{6}$
$6\beta =5$