<s>[INST] Evaluation of $\lim\limits_{x\rightarrow0} \frac{\tan(x)-x}{x^3}$ One of the previous posts made me think of the following question: Is it possible to evaluate this limit without L'Hopital and Taylor?
$$\lim_{x\rightarrow0} \frac{\tan(x)-x}{x^3}$$
 [/INST] Encouraged by Hans Lundmark's answer, I'm posting my own solution without derivatives and integrals.
The triple-angle formula for $\tan$ is $$\tan 3\theta = \frac{3\tan\theta-\tan^3\theta}{1-3\tan^2\theta}.$$
Suppose $\lim_{x\to0}(\tan x-x)/x^3 = c$. Letting $x = 3\theta$, we then have
$$\begin{align}
c &= \lim_{x\to0} \frac{\tan x-x}{x^3} \\
&= \lim_{\theta\to0} \frac{\tan 3\theta-3\theta}{27\theta^3} \\
&= \lim_{\theta\to0} \frac{3\tan\theta - \tan^3\theta-3\theta+9\theta\tan^2\theta}{27\theta^3(1-3\tan^2\theta)}
\end{align}$$
We can get rid of $1/(1-3\tan^2\theta)$ because its limit is $1$. Next we start pulling out terms and find
$$\begin{align}
c &= \lim_{\theta\to0}\frac{3\tan\theta-3\theta}{27\theta^3} - \lim_{\theta\to0}\frac{\tan^3\theta}{27\theta^3} + \lim_{\theta\to0}\frac{9\theta\tan^2\theta}{27\theta^3} \\
&= \frac19c - \frac1{27} + \frac{1}{3},
\end{align}$$
because $\lim_{\theta\to0}(\tan\theta)/\theta = 1$. So $8c/9 = 8/27$, or $c = 1/3$.
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