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 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}
