limx tends to 0 sin^2x(1-cos square x) whole divided by x^6

$\underset{x\to 0}{\lim } \frac{{\sin }^{2}x\left(1-{\cos }^{2}x\right)}{x^6}=\underset{x\to 0}{\lim } \frac{{\sin }^{4}x}{x^6}$

This is of the $\frac{0}{0}$ form. Keep on applying L'Hospitals Rule until the $\frac{0}{0}$ form is no more encountered.

$\underset{x\to 0}{\lim }\frac{{\sin }^{4}x}{x^6}=\underset{x\to 0}{\lim } \frac{4{\sin }^{3}x\cos x}{6x^5}$
$=\frac{2}{3}.\underset{x\to 0}{\lim } \frac{3{\sin }^{3}x\cos x+{\sin }^{3}x\left(-\sin x\right)}{5x^4}$
$=\frac{2}{3}.\underset{x\to 0}{\lim } \frac{3{\sin }^{2}x(-\sin x)+6\sin x{\cos }^{2}x-4{\sin }^{3}x\cos x}{20x^3}$
$=\frac{2}{3}.\underset{x\to 0}{\lim } \frac{-9{\sin }^{2}x\cos x+6(-2\cos x\sin x)+6{\cos }^{3}x-12{\sin }^{2}x\cos x}{60x^2}$
$=\frac{2}{3}.\underset{x\to 0}{\lim } \frac{-33{\sin }^{2}x\cos x+6{\cos }^{3}x}{60x^2}$
$=\frac{2}{3}.\underset{x\to 0}{\lim } \frac{-33\left\{2\sin x{\cos }^{2}x+{\sin }^{2}x\left(-\cos x\right)-18{\cos }^{2}x\sin x\right\}}{120x}$
$=\frac{2}{3}.\underset{x\to 0}{\lim } \frac{33{\sin }^{2}x\cos x-84\sin x\cos x^2}{120x}$
$=\frac{2}{3}.\underset{x\to 0}{\lim } \frac{33\left\{2\sin x{\cos }^{2}x-{\sin }^{3}x\right\}-84\left\{{\cos }^{3}x-2{\sin }^{2}x\cos x\right\}}{120}$
$=\frac{2}{3}\times \frac{33\left\{2\times 0-0\right\}-84\left\{1-0\right\}}{120}=-\frac{2}{3}\times \frac{84}{120}=-\frac{13}{30}$