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.

The limit $$\lim_{x\to 0}\left(\frac {\sin x}x \cdot \frac 1{x^2} - \frac 1{x^2}\right)$$ is of the indeterminate form $\infty - \infty$ so
$$\lim_{x\to 0}\left(\frac {\sin x}x \cdot \frac 1{x^2} - \frac 1{x^2}\right) \neq \lim_{x\to 0}\left(\frac {\sin x}x \cdot \frac 1{x^2}\right) - \lim_{x\to 0}\left(\frac 1{x^2}\right)$$ because the RHS makes no sense.
