What's the mistake I'm making in evaluating the below limit? $\lim_{x \to 0}\frac{x\sin(\sin x) - \sin^{2}x}{x^{6}}$ Below is my solution for this limit. I get $0$, but the answer, is in fact, $\frac{1}{18}$. What's the mistake I'm making?
{From here: A limit problem $\lim\limits_{x \to 0}\frac{x\sin(\sin x) - \sin^{2}x}{x^{6}}$}

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

$$ \sin t = t - \frac{t^3}{6} + \frac{t^5}{120} \mp ...  $$
$$ \sin \sin x = \sin x -\frac{\sin^3 x}{6} + \frac{\sin^5 x}{120} \mp  $$
$$ \sin \sin x = \left( x - \frac{x^3}{6} + \frac{x^5}{120} \mp ... \right) - \frac{\left( x - \frac{x^3}{6} \pm \right)^3}{6} + \frac{\left( x - \mp \right)^5}{120} \pm $$
$$ \sin \sin x = \left( x - \frac{x^3}{6} + \frac{x^5}{120} \mp ... \right) - \frac{\left( x^3 - \frac{x^5}{2} \pm \right)}{6} + \frac{ x^5 - \mp }{120} \pm $$
$$ \sin \sin x =  x - \frac{x^3}{3} + \frac{x^5}{10} \mp  $$
$$ x \sin \sin x =  x^2 - \frac{x^4}{3} + \frac{x^6}{10} \mp  $$
$$  \sin^2 x =  x^2 - \frac{x^4}{3} + \frac{2 x^6}{45} \mp  $$
$$ x \sin \sin x - \sin^2 x =   \frac{x^6}{10} -  \frac{2 x^6}{45} \pm  $$
$$ x \sin \sin x - \sin^2 x =   \frac{9x^6}{90} -  \frac{4 x^6}{90} \pm  $$
$$ x \sin \sin x - \sin^2 x =   \frac{5x^6}{90} \pm  $$
$$ x \sin \sin x - \sin^2 x =   \frac{x^6}{18} + O( x^7)  $$
Meanwhile, the function depicted is even, all exponents in the final expression must be even, so we can say the slightly stronger
$$ x \sin \sin x - \sin^2 x =   \frac{x^6}{18} + O( x^8)  $$
