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.
 A: Famously, a sector with a small angle proves $\lim_{x\to0}\tfrac{\sin x}{x}=1$ and hence$$\lim_{x\to0}\tfrac{1-\cos x}{x^2}=\lim_{x\to0}\tfrac{\sin^2\tfrac{x}{2}}{2(x/2)^2}=\tfrac12.$$We can improve the former result: if you approximate a sector's arc as a parabola, you can prove$$\lim_{x\to0}\frac{1-\tfrac{\sin x}{x}}{x^2}=\tfrac16.$$Finally,$$\lim_{x\to0}\frac{\tfrac{\tan x}{x}-1}{x^2}=\lim_{x\to0}\frac{\tfrac{\sin x}{x}-\cos x}{x^2}=\lim_{x\to0}\frac{1-\cos x}{x^2}-\lim_{x\to0}\frac{1-\tfrac{\sin x}{x}}{x^2}=\tfrac12-\tfrac16=\tfrac13.$$
A: Look at the trigonometric circle and approximate the chord $s$ with the angle $\theta$.

Here $x=\cos \theta$ and $y=\sin \theta$.
From the big triangle, you have $x^2+y^2=1$ which is obvious, and from the small circle, you have $s^2 = y^2 + (1-x)^2$. But with approximation $s \approx \theta$ the above is
$$\theta^2 \approx y^2 + (1-x)^2 \tag{1}$$
Now sub $y=\sqrt{1-x^2}$ to get the well-known approximation
$$ \theta^2 \approx 1-x^2 + (1-x)^2 = 2 (1-x) $$
or
$$ x = \cos \theta \approx 1 - \tfrac{1}{2} \theta^2 \tag{2}$$
Now use (2) in (1) to get $\theta^2 \approx y^2 + \tfrac{1}{4} \theta^4$, but since we are keeping this a second order approximation $\theta^4 \approx 0$ and
$$ y = \sin \theta \approx \theta \tag{3} $$
Finally, $\tan \theta = \tfrac{y}{x} \approx \tfrac{\theta}{1-\tfrac{1}{2} \theta^2}$. But use the property $\tfrac{1}{1-z} = 1 + \frac{1}{1-z} z \approx 1 + z$ with $z=\tfrac{1}{2} \theta^2$ to get the final approximation.
$$ \tfrac{y}{x} = \tan \theta \approx \theta \left( 1 + \tfrac{1}{2} \theta^2 \right) \tag{4a} $$
But note that the last one is based on two approximations.

Another approach with $\tan$ would be to approximate once $\tan \theta = \tfrac{y}{x} = \theta \frac{\sqrt{1-\tfrac{1}{2}z}}{1-z}$ with $z=\tfrac{1}{2}\theta^2$ and the approximation $ \tan \theta \approx \theta \left( 1 + \tfrac{3}{4} z \right)$
$$ \tfrac{y}{x} = \tan \theta \approx \theta \left( 1 + \tfrac{3}{8} \theta^2 \right) \tag{4b} $$
For example with $\theta=0.6$ you have $\tan(0.6) = 0.68413681$, $0.6 \left( 1 + \tfrac{1}{2} 0.6^2 \right)=0.708$ an 3.5% error, and $0.6 \left( 1 + \tfrac{3}{8} 0.6^2 \right)=0.681$ an 0.45% error.
So (4b) is at least an order of magnitude better at approximating $\tan \theta$ than (4a). A graphical comparison of the error $f(x)/\tan x-1$ is shown below:
 
A: 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 $\frac1{1-x}=1+x+x^2+\dots\approx1+x$,
$$
\begin{align}
\tan(x)
&=\frac{\sin(x)}{\cos(x)}\tag{6a}\\
&\approx\left(x-\frac{x^3}6\right)\left(1+\frac{x^2}2\right)\tag{6b}\\
&\approx x+\frac{x^3}3\tag{6c}
\end{align}
$$
