Why does $\lim_{x \to 0}\frac{\tan x -x}{x^3}=\lim_{x \to 0}\frac{\tan 2x -2x}{8x^3}$?

Question

In a post about evaluating limits without L'Hopital's Rule or series expansion, one of the limits used as an example was this:

$$ \lim_{x \to 0}\frac{\tan x -x}{x^3} $$

This expression was said to be equal to this:

$$ \lim_{x \to 0}\frac{\tan 2x -2x}{8x^3} $$

I don't understand how this follows. I tried using $$\tan 2x\equiv\frac{2 \tan x}{1-\tan ^2x}$$ but it didn't seem to work. How can the two limits be shown to be equal to each other?

Answer 1

It's a special case of $\lim_{x\to0}f(x)=\lim_{x\to0}f(2x)$.

Answer 2

The result follows considering $y=2x \to 0$ and more in general for any $f(x) \to 0$, not identically equal to $0$, we have

$$\lim_{x \to 0}\frac{\tan x -x}{x^3}=\lim_{x \to 0}\frac{\tan (f(x)) -f(x)}{(f(x))^3}$$