$\begin{aligned} & \lim _{x \rightarrow 0} \frac{\tan ^3 x-\sin ^3 x}{x^5} \\ & =\lim _{x \rightarrow 0} \frac{\left[\begin{array}{c}\left(x+\frac{x^3}{3}+\frac{2}{15} x^5+\ldots\right)^3 \\ -\left(x-\frac{x^3}{3 !}+\frac{x^5}{5 !}+\ldots\right)^3\end{array}\right]}{x^5} \\ & =\lim _{x \rightarrow 0} \frac{\left[\begin{array}{c}\left(1+\frac{x^2}{2}+\frac{2}{15} x^4+\ldots\right)^3 \\ -\left(1-\frac{x^2}{3 !}+\frac{x^4}{5 !}+\ldots\right)^3\end{array}\right]}{x^2} \\ & \left\{\left(1+\frac{x^2}{3}+\frac{2}{15} x^4+\ldots\right)-\left(1-\frac{x^2}{3 !}+\frac{x^4}{5 !}+\ldots\right)\right\} \\ & =\lim _{x \rightarrow 0} \frac{\left\{\begin{array}{c}\left(1+\frac{x^2}{3}+\ldots\right)^2+\left(1-\frac{x^2}{3 !}+\ldots\right)^2 \\ +\left(1+\frac{x^2}{3}+\ldots\right)\left(1-\frac{x^2}{3 !}+\ldots\right)\end{array}\right\}}{x^2} \\ & \left.\text { [using }\left(a^3-b^3\right)=(a-b)\left(a^2+b^2+a b\right)\right] \\ & =\lim _{x \rightarrow 0} \frac{\left(\frac{x^2}{2}+\frac{x^4}{8}+\ldots\right)\left\{\begin{array}{c}\left.\left(1+\frac{x^2}{3}+\ldots\right)^2+\left(1-\frac{x^2}{3 !}+\ldots\right)^2\right\} \\ +\left(1+\frac{x^2}{3}+\ldots\right)\left(1-\frac{x^2}{3}+\ldots\right)\end{array}\right\}}{x^2} \\ & =\lim _{x \rightarrow 0}\left(\frac{1}{2}+\frac{x^2}{8}\right)\left\{\begin{array}{c}\left(1+\frac{x^2}{3}+\ldots\right)^2+\left(1-\frac{x^2}{3 !}+\ldots\right)^2 \\ +\left(1+\frac{x^2}{3}+\ldots\right)\left(1-\frac{x^2}{3 !}+\ldots\right)\end{array}\right\} \\ & \end{aligned}$
$\begin{aligned} & =\left(\frac{1}{2}+0\right)\left\{(1+0+\ldots)^2+(1-0+\ldots)^2\right. \\ & =\frac{1}{2}(1+1+1)=\frac{3}{2}\end{aligned}$