Limit, solution in unusual way I have a problem with solution of this limit:
$$\lim_{x\to 0}{\frac{\tan{x}-x}{x^2}}$$
Of course, it's a very easy to solve, using (twice) L'Hôpital's rule, but I need to find out, how to do this without this rule.
I stuck in this point:
$$\lim_{x\to 0}{\frac{\frac{\sin^2{x}}{x^2}\cdot\frac{1}{\cos^2{x}} - 1}{2x}}$$
Everything I need to know is how to eliminate $\frac{\sin^2{x}}{x^2}$, because - as my tutor said - I can't simply substitute $1$ for this expression.
Thanks for help.
PS: It's not a homework. My tutor showed this problem as a puzzle and said, that it would be a good exercise to solve this without L'Hôpital's rule.
EDIT: Here is a way I got to the point $\lim_{x\to 0}{\frac{\frac{\sin^2{x}}{x^2}\cdot\frac{1}{\cos^2{x}} - 1}{2x}}$:
\begin{align*}
\lim_{x->0}{\frac{\tan{x}-x}{x^2}} &= \lim_{x->0}{\frac{(\frac{\sin{x}}{\sqrt{1-\sin^2{x}}}-x)\cdot (\frac{\sin{x}}{\sqrt{1-\sin^2{x}}}+x)}{x^2(\frac{\sin{x}}{\sqrt{1-\sin^2{x}}}+x)}}\\
 &= \lim_{x->0}{\frac{\frac{\sin^2{x}}{x^2\cos^2{x}}-1}{x(\frac{\sin{x}}{x}\frac{1}{\cos{x}}+1)}}= \lim_{x\to 0}{\frac{\frac{\sin^2{x}}{x^2}\cdot\frac{1}{\cos^2{x}} - 1}{2x}}
\end{align*}

Here is an attempt at a geometric proof.

(Figure thanks to J.M)
Consider $\triangle ABC$ such that $\angle{BCA} = x$. Let $BC=1$ and so $AB = \tan x$.
Let $BE$ be the arc of radius 1 and angle $x$ drawn with $C$ as the center (note that $E$ is on AC, between $A$ and $D$ and is kind of hidden in the brown region). Note that $CE = 1$.
Now the area of the gray region is $\dfrac{\tan x}{2} - \dfrac{x}{2}$ (area of $\triangle ABC$ - area of the sector $CBE$).
Let $D$ be the perpendicular on the hypotenuse $AC$ from $B$. It can be seen that $CD = \cos x$ and thus distance from $D$ to $C$ is less than distance from $E$ to $C$ (which is $1$).
Thus the area of the gray region is less than the area of $\triangle BAD$ (gray + brown).
Now $AD = \dfrac{\sin^2 x}{\cos x}$ and thus we have that
$$ 0 < \dfrac{\tan x - x}{2} \lt \dfrac{\sin^3 x}{2\cos x}$$
And so 
$$ 0 < \dfrac{\tan x - x}{2x^2} \lt \dfrac{\sin^3 x}{2x^2 \cos x}$$
Since we know that $\lim_{x \to 0+} \dfrac{\sin x}{x} = 1$, and that $\dfrac{\tan x - x}{x^2}$ is an odd function, that the limit is $0$, follows.

Previous answer, which was a feeble attempt at being pedantic:
For a way to find the limit without using more advanced concepts like McLaurin series etc...
Consider
$$\dfrac{\tan(2x) - 2x}{(2x)^2} = \dfrac{ \frac{2\tan x}{1-\tan^2 x} - 2x}{4x^2}$$
$$ = \dfrac{(2\tan x - 2x) + 2x \tan^2 x}{4x^2(1 - \tan^2 x)} = (\dfrac{\tan x - x}{2x^2} +  \dfrac{x\tan^2 x}{2x^2}) \dfrac{1}{1-\tan^2 x}$$
Therefore, taking limits as $\displaystyle x \to 0$
$$ L = (\dfrac{L}{2} + 0)\dfrac{1}{1-0}$$
Thus
$$L = 0$$
There is one problem with the above, though. Can you tell what that is?
(Or rather more simply, replace $\displaystyle x$ with $\displaystyle -x$)
