calculus - $lim_{x to 0} {1-xcot xover x^2}$

What's the value of this limit? I keep getting half, but the answer's 1/3. I think there might be a problem with writing ${x\cot x \over x^2 }$ as $ {\cos x \over x^2}$. Is there ?
I get (1-cosx)/x^2 on using this substitution, and if I use L'hôpital's Rule on this, I get the answer as sinx/2x = 1/2.

Edit: I can only use results like $\sin x=x$ and L'Hospital's rule. No series expansion.

Edit 2: Here's my attempt:

1/x^2 - cosx/(x*sinx) = 1/x^2 - (cosx/x^2)(x/sinx) = 1/x^2 - (cosx/x^2)(1)
{Since sinx/x=1}
= (1-cosx)/x^2
Being a 0/0 case, L'hôpital's Rule:
sinx/2x = (1/2)(sinx/x) = 1/2, since sinx/x=1

Answer

Rewrite the limit as
$$
\lim_{x\to0}\frac{\sin x-x\cos x}{x^2\sin x}=\lim_{x\to0}\frac{\sin x-x\cos x}{x^3}\frac{x}{\sin x}

$$
The second fraction has limit $1$, so you just need to compute
$$
\lim_{x\to0}\frac{\sin x-x\cos x}{x^3}=\lim_{x\to0}\frac{\cos x-\cos x+x\sin x}{3x^2}=\lim_{x\to0}\frac{1}{3}\frac{\sin x}{x}
$$

Your attempt is faulty, because
$$
\lim_{x\to0}\frac{1}{x^2}=\infty,\qquad
\lim_{x\to0}\frac{\cot x}{x}=\infty

$$
so you have a form $\infty-\infty$ and you can't use that $\sin x=x$ (which is false, by the way, avoid thinking like that).