Question

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

Answer 1

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