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?
$$\lim_{x\rightarrow0} \frac{\tan(x)-x}{x^3}$$

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.
