<s>[INST] Elementary proof (involving no series expansion) that $\tan x \approx x+\frac{x^3}{3}$ for small $x$ For small values of $x$, the following widely used approximations follow immediately by Taylor expansion:

*

*$\sin x \approx x$

*$\cos x \approx 1-\frac{x^2}{2}$

*$\tan x \approx x +\frac{x^3}{3}$
I am looking for a justification of these approximations without the use of series expansions.
By purely geometric considerations, it is easy to see that for small values of $x$, we have
$$ \sin x < x < \tan x.$$
Division by $\cos x$ and an application of the squeeze lemma yield
$$\frac{\sin x}{x}\xrightarrow{x\to0}1$$
and hence the approximation (1.). Using the identity $\cos x = 1-2 \sin^2 \frac{x}{2}$, the approximation (2.) also follows.
Can one justify the approximation (3.) by a similarly elementary argument without using Taylor expansion?
I tried around using the angle addition theorems, but I did not really get anywhere, mainly because I could not make the factor $\frac13$ appear anywhere.
 [/INST] Proof that $\boldsymbol{\lim\limits_{x\to0}\frac{x-\sin(x)}{x^3}}=\frac16$
As shown in $(2)$ from this answer, since $\cos(x)$ is continuous, given any $\epsilon\gt0$, we can find a $\delta\gt0$ so that if $0\lt|x|\le\delta$, we have
$$
1-\epsilon\le\cos(x)\le\frac{\sin(x)}x\le1\tag1
$$
Thus, assuming $0\lt|x|\le\delta$,
$$
\begin{align}
\frac{x-\sin(x)}{x^3}
&=\frac1{x^3}\sum_{n=0}^\infty\left(2^{n+1}\sin\left(x/2^{n+1}\right)-2^n\sin\left(x/2^n\right)\right)\tag{2a}\\
&=\frac1{x^3}\sum_{n=0}^\infty\left(2^{n+1}\sin\left(x/2^{n+1}\right)-2^{n+1}\sin\left(x/2^{n+1}\right)\cos\left(x/2^{n+1}\right)\right)\tag{2b}\\
&=\frac1{x^3}\sum_{n=0}^\infty2^{n+1}\sin\left(x/2^{n+1}\right)\left(1-\cos\left(x/2^{n+1}\right)\right)\tag{2c}\\
&=\frac1{x^3}\sum_{n=0}^\infty\frac{2^{n+1}\sin^3\left(x/2^{n+1}\right)}{1+\cos\left(x/2^{n+1}\right)}\tag{2d}\\
&=\frac12\sum_{n=0}^\infty\frac1{2^{2n+2}}\left[(1-\epsilon)^3,\frac1{1-\epsilon/2}\right]\tag{2e}\\
&=\frac16\left[(1-\epsilon)^3,\frac1{1-\epsilon/2}\right]\tag{2f}
\end{align}
$$
Explanation:
$\text{(2a)}$: apply $\lim\limits_{x\to0}\frac{\sin(x)}x=1$ and the telescoping sum
$\text{(2b)}$: $\sin(2x)=2\sin(x)\cos(x)$
$\text{(2c)}$: factor
$\text{(2d)}$: $\sin^2(x)=(1-\cos(x))(1+\cos(x))$
$\text{(2e)}$: apply $(1)$ with $[a,b]$ representing a number in that range
$\text{(2f)}$: evaluate the sum
Therefore, by $(2)$ and the Squeeze Theorem,
$$
\lim_{x\to0}\frac{x-\sin(x)}{x^3}=\frac16\tag3
$$

The Approximations
The limit $(3)$ gives the following approximation
$$
\sin(x)\approx x-\frac{x^3}6\tag4
$$
Since the Binomial Theorem gives $(1-x)^2=1-2x+x^2\approx1-2x$, we have $\sqrt{1-x}\approx1-\frac x2$.
$$
\begin{align}
\cos(x)
&=\sqrt{1-\sin^2(x)}\tag{5a}\\
&\approx\sqrt{1-x^2}\tag{5b}\\
&\approx1-\frac{x^2}2\tag{5c}
\end{align}
$$
Finally, since the Geometric Series gives $\frac1{1-x}=1+x+x^2+\dots\approx1+x$,
$$
\begin{align}
\tan(x)
&=\frac{\sin(x)}{\cos(x)}\tag{6a}\\
&\approx\left(x-\frac{x^3}6\right)\left(1+\frac{x^2}2\right)\tag{6b}\\
&\approx x+\frac{x^3}3\tag{6c}
\end{align}
$$
</s>