Computing $\lim\limits_{x \to 0} {\frac{\sin x-x}{x^3}}$ So I was solving the Cengage Mathematics Calculus book for JEE Adv. and I came across this question in the examples $$\lim\limits_{x \to 0} {\frac{\sin x-x}{x^3}}$$
The solution given in the book uses the expansion of $\sin x$ in solving the question, whereas WolframAlpha used L'Hopital's Rule in solving it.
My first instinct to solving the question when I read it was like this-
$$ \lim\limits_{x \to 0} {\frac{\sin x-x}{x^3}} $$
$$ \implies \lim\limits_{x \to 0} \frac{\sin x}{x} \times \frac{1}{x^2} - \frac{x}{x^3} $$
$$ \implies \lim\limits_{x \to 0} \frac{\sin x}{x} \times \lim\limits_{x \to 0} \frac{1}{x^2} - \lim\limits_{x \to 0} \frac{x}{x^3} $$
$$ \implies 1 \times \lim\limits_{x \to 0} \frac{1}{x^2} - \lim\limits_{x \to 0} \frac{1}{x^2} $$
$$ \implies 1 \times \lim\limits_{x \to 0} \frac{1}{x^2} - \frac{1}{x^2} $$
$$ \implies 1 \times 0 $$
$$ \implies 0 $$
I do not understand, where I am going wrong with this procedure. I would really appreciate it if someone can answer.
Thank you.
For reference here is the solution given in Cengage.

Assuming the limit exists, let
$$l=\lim_{x \to 0}\frac{\sin x-x}{x^3} $$
$$ l=\lim_{x \to 0}\frac{\sin x-x}{x^3}\overset{x=2t}{=}\lim_{t \to 0}\frac{\sin 2t-2t}{8t^3}=\lim_{x \to 0}\frac{2\sin x \cos x-2x}{8x^3} $$
Hence
$$4l=\lim_{x \to 0}\frac{\sin x \cos x-x}{x^3} $$
Now consider
$$3l=4l-l=\lim_{x \to 0}\frac{\sin x \cos x-x}{x^3}-\lim_{x \to 0}\frac{\sin x-x}{x^3}=\lim_{x \to 0}\frac{\sin x \cos x-\sin x}{x^3}\\ 3l=\lim_{x \to 0}\frac{\ \sin x}{x} \lim_{x \to 0}\frac{\cos x-1}{x^2}=-\frac{1}{2} $$
Finally
$$ \boxed{l=\lim_{x \to 0}\frac{\sin x-x}{x^3}=-\frac{1}{6}}$$
