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?

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

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

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  • $\begingroup$ 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? $\endgroup$ – Joe Sep 10 at 19:12
  • $\begingroup$ I think you should put the question in comment in place of the actual question $\endgroup$ – Buraian Sep 10 at 19:13
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    $\begingroup$ @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 $0$. $\endgroup$ – 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

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

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