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?
$$\lim_{x\rightarrow0} \frac{\tan(x)-x}{x^3}$$
 A: There's the Cauchy or extended mean value theorem which says that ${\displaystyle {f(x) - f(y) \over g(x) - g(y)} = {f'(c) \over g'(c)}}$ for some $c$ between $x$ and $y$. You can apply it here with $f(x) = \tan(x) - x$ and $g(x) = x^3$, and you get that for some $0 < c < x$ you have
$$\frac{\tan(x) - x}{x^3} = {\sec^2(c) - 1 \over 3c^2}$$
$$= {\tan^2(c) \over 3c^2}$$
$$ = {1 \over 3}{1 \over \cos^2(c)}{\sin^2(c) \over c^2}$$
Now take limits as $x$ goes to zero; $c$ goes to zero and the limit is $1/3$.
This is all somewhat tongue in cheek of course, since you can get L'Hopital pretty quickly from the extended mean value theorem, but it does satisfy your request to not use it or Taylor polynomials :)
A: 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 $c = 1/3$.
A: Here is a different approach. Let $$L = \lim_{x \to 0} \dfrac{\tan(x) - x}{x^3}$$
Replacing $x$ by $2y$, we get that
\begin{align}
L & = \lim_{y \to 0} \dfrac{\tan(2y) - 2y}{(2y)^3} = \lim_{y \to 0} \dfrac{\dfrac{2 \tan(y)}{1 - \tan^2(y)} - 2y}{(2y)^3}\\
& = \lim_{y \to 0} \dfrac{\dfrac{2 \tan(y)}{1 - \tan^2(y)} - 2 \tan(y) + 2 \tan(y) - 2y}{(2y)^3}\\
& = \lim_{y \to 0} \dfrac{\dfrac{2 \tan^3(y)}{1 - \tan^2(y)} + 2 \tan(y) - 2y}{(2y)^3}\\
& = \lim_{y \to 0} \left(\dfrac{2 \tan^3(y)}{8y^3(1 - \tan^2(y))} + \dfrac{2 \tan(y) - 2y}{8y^3} \right)\\
& = \lim_{y \to 0} \left(\dfrac{2 \tan^3(y)}{8y^3(1 - \tan^2(y))} \right) + \lim_{y \to 0}  \left(\dfrac{2 \tan(y) - 2y}{8y^3} \right)\\
& = \dfrac14 \lim_{y \to 0} \left(\dfrac{\tan^3(y)}{y^3} \dfrac1{1 - \tan^2(y)} \right) + \dfrac14 \lim_{y \to 0}  \left(\dfrac{\tan(y) - y}{y^3} \right)\\
& = \dfrac14 + \dfrac{L}4
\end{align}
Hence, $$\dfrac{3L}{4} = \dfrac14 \implies L = \dfrac13$$
EDIT
In Hans Lundmark answer, evaluating the desired limit boils down to evaluating $$S=\lim_{x \to 0} \dfrac{\sin(x)-x}{x^3}$$ The same idea as above can be used to evaluate $S$ as well.
Replacing $x$ by $2y$, we get that \begin{align}
S & = \lim_{y \to 0} \dfrac{\sin(2y) - 2y}{(2y)^3} = \lim_{y \to 0} \dfrac{2 \sin(y) \cos(y) - 2y}{8y^3}\\
& = \lim_{y \to 0} \dfrac{2 \sin(y) \cos(y) - 2 \sin(y) + 2 \sin(y) - 2y}{8y^3}\\
& = \lim_{y \to 0} \dfrac{2 \sin(y) - 2y}{8y^3} + \lim_{y \to 0} \dfrac{2 \sin(y) \cos(y)-2 \sin(y)}{8y^3}\\
& = \dfrac14 \lim_{y \to 0} \dfrac{\sin(y) - y}{y^3} - \dfrac14 \lim_{y \to 0} \dfrac{\sin(y) (1 - \cos(y))}{y^3}\\
& = \dfrac{S}4 - \dfrac14 \lim_{y \to 0} \dfrac{\sin(y) 2 \sin^2(y/2)}{y^3}\\
& = \dfrac{S}4 - \dfrac18 \lim_{y \to 0} \dfrac{\sin(y)}{y} \dfrac{\sin^2(y/2)}{(y/2)^2}\\
& = \dfrac{S}4 - \dfrac18 \lim_{y \to 0} \dfrac{\sin(y)}{y} \lim_{y \to 0} \dfrac{\sin^2(y/2)}{(y/2)^2}\\
& = \dfrac{S}4 - \dfrac18\\
\dfrac{3S}4 & = - \dfrac18\\
S & = - \dfrac16
\end{align}
A: The statement $\dfrac{\tan(x)-x}{x^3} \to c$ as $x \to 0$ is equivalent to 
$\tan(x) = x + c x^3 + o(x^3)$ as $x \to 0$, so this is a statement about a
Taylor polynomial of $\tan(x)$, and I'm not sure what would count as doing
that "without Taylor".  However, one thing you could do is start from $$\sin(x) = x + o(x)$$ integrate to get $$\cos(x) = 1 - x^2/2 + o(x^2)$$ then $$\sec(x) = \frac{1}{1-x^2/2 + o(x^2)} = 1 + x^2/2 + o(x^2)$$ $$\sec^2(x) = \left(1 + x^2/2 + o(x^2)\right)^2 = 1 + x^2 + o(x^2)$$ and integrate again to get
$$\tan(x) = x + x^3/3 + o(x^3)$$
A: Here's a way which avoids derivatives and integrals.
Assume that we know that $\frac{\sin x}{x} \to 1$ as $x \to 0$.
Then we also know that
$\frac{1-\cos x}{x^2} = \frac12 \left( \frac{\sin(x/2)}{x/2} \right)^2 \to \frac12$.
Now,
$$
\frac{\tan x - x}{x^3} = \frac{1}{\cos x} \left( \frac{\sin x - x}{x^3} + \frac{1-\cos x}{x^2} \right),
$$
so we are done if we can compute $\lim_{x \to 0} \frac{\sin x - x}{x^3} = -\frac16$.
The reason that I rewrote it like this is that I was asked by a
colleague about ten years ago whether that limit could be done in an
elementary way. :-) I came up with the following:
Let
$$
f(x) = \frac{x - \sin x}{x^3} = \frac{1 - \frac{\sin x}{x}}{x^2}.
$$
(Here I've changed the sign so that the limit
will be positive.)
Since $f$ is an even function, it's enough to consider $x>0$.
Fix a positive integer $n$. To begin with, we have
$$
x  =  2^n \frac{x}{2^n}  >  2^n \sin \frac{x}{2^n}.
$$
(I'm assuming that we also know that $0 < \sin x < x < \tan x$ for $0 < x < \pi/2$.)
Multiply this inequality by $\prod_{k=1}^n \cos\frac{x}{2^k}$ and use the double angle formula repeatedly,
as follows (illustrated for the case $n=3$):
$$
\begin{split}
  x \cos\frac{x}{8} \cos\frac{x}{4} \cos\frac{x}{2}
  & > 2^3 \sin\frac{x}{8} \cos\frac{x}{8} \cos\frac{x}{4} \cos\frac{x}{2}
  \\
  & = 2^2 \sin\frac{x}{4} \cos\frac{x}{4} \cos\frac{x}{2}
  \\
  & = 2^1 \sin\frac{x}{2} \cos\frac{x}{2}
  \\
  & = \sin x.
\end{split}
$$
This implies (again for $n=3$, but the general pattern is hopefully clear)
$$
\begin{split}
  1 - \frac{\sin x}{x}
  & > 1 - \cos\frac{x}{8} \cos\frac{x}{4} \cos\frac{x}{2}
  \\
  & =
  \left( 1 - \cos\frac{x}{8} \right)
  + \cos\frac{x}{8} \left( 1 - \cos\frac{x}{4} \right)
  + \cos\frac{x}{8} \cos\frac{x}{4} \left( 1 - \cos\frac{x}{2} \right).
\end{split}
$$
We know that $\frac{1 - \cos(x/2^k)}{x^2} = \frac{1 - \cos(x/2^k)}{2^{2k} (x/2^k)^2} \to \frac{1}{2^{2k+1}}$,
so after dividing this inequality by $x^2$ we find in the limit
(for general $n$) that
$$
\liminf_{x \to 0^+} f(x) \ge \sum_{k=1}^n \frac{1}{2^{2k+1}} = \frac16 \left( 1 - \frac{1}{4^n} \right).
$$
This holds for every $n$, hence
$$
\liminf_{x \to 0^+} f(x) \ge \frac16.
$$
The other direction is similar.
Start with
$$
x = 2^n \frac{x}{2^n} < 2^n \tan\frac{x}{2^n} = 2^n \frac{\sin(x/2^n)}{\cos(x/2^n)}.
$$
This leads to
$$
\begin{split}
  1 - \frac{\sin x}{x}
  & < 1 - \cos\frac{x}{2^n} \cdot (\text{same product of cosines as above})
  \\
  & = 1 - \cos\frac{x}{2^n} + \cos\frac{x}{2^n} \cdot (1 - (\text{that product}))
  \\
  & = 1 - \cos\frac{x}{2^n} + \cos\frac{x}{2^n} \cdot (\text{same expression as above}).
\end{split}
$$
Divide by $x^2$ and let $x \to 0^+$:
$$
\limsup_{x \to 0^+} f(x) \le \frac{1}{2^{2n+1}} + \frac16 \left( 1 - \frac{1}{4^n} \right).
$$
Let $n \to \infty$:
$$
\limsup_{x \to 0^+} f(x) \le \frac16.
$$
It follows that $\lim_{x \to 0^+} f(x) = \frac16$, and therefore by symmetry $\lim_{x \to 0} f(x) = \frac16$,which is what we wanted to show.
