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

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?

• Substitute $y=2x$ in the first limit, and then, replace $y$ by $x$. – TheSilverDoe Sep 10 at 19:09

It's a special case of $$\lim_{x\to0}f(x)=\lim_{x\to0}f(2x)$$.
• Thank you very much. Is it true that, in general, $\lim_{x \to 0}f(kx)=\lim_{x \to 0}f(x)$, where $k$ is a constant? – Joe Sep 10 at 19:12
• @Joe If the limit exists & $k\ne0$, yes. There's an easy $\epsilon$-$\delta$ proof. The case $k=0$ exists iff $f$ is continuous at $0$. – J.G. Sep 10 at 19:15
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
$$\lim_{x \to 0}\frac{\tan x -x}{x^3}=\lim_{x \to 0}\frac{\tan (f(x)) -f(x)}{(f(x))^3}$$