limit of $\lim_{x\to 0} \frac{(\sin x)^2 - (x \cos x) ^ 2}{x^4}$ I have question. I want to solve this limit. it's $\frac{0}{0}$ so we have to change it. there is two way with two different value.
$\lim_{x\to 0} \frac{(\sin x)^2 - (x \cos x) ^ 2}{x^4}$  
First way:
before that we know that $\lim_{x\to 0} \frac{\sin x}{x}$ or $\lim_{x\to 0} \frac{(\sin x) ^ 2}{x ^ 2}$ is equal to 1 so:
$\lim_{x\to 0} \frac{(\sin x)^2 - (x \cos x) ^ 2}{x^4} = \lim_{x\to 0} \frac{x^2 - x ^ 2 (\cos x) ^ 2}{x^4} = \lim_{x\to 0} \frac{x^2 (1 - (\cos x) ^ 2)}{x^4} = \lim_{x\to 0} \frac{(\sin x) ^ 2}{x^2} = 1$  
second way:
before that we know that $\sin x \sim x - \frac{x^3}{6}$ and $\cos x \sim 1 - \frac{x^2}{2}$ so:
$\lim_{x\to 0} \frac{(\sin x)^2 - (x \cos x) ^ 2}{x^4} = \lim_{x\to 0} \frac{(\sin x) - (x \cos x)}{x^3} * \frac{(\sin x) + (x \cos x)}{x} = \lim_{x\to 0} \frac{x - \frac{x^3}{6} - x + \frac{x^3}{2}}{x^3} * \frac{x - \frac{x^3}{6} + x - \frac{x^3}{2}}{x} = (\frac{1}{2} - \frac{1}{6}) * 2 = \frac{2}{3}$    
Update:
Is it possible to explain more? We have limit and we solve like this and that's work but in this limit we can't use $\lim_{x\to0}\left(\frac{\sin x}x\right)^2=1$. 
this in another limit:
$\lim_{x\to0} \frac{1 - \cos 2x}{x^2}= \lim_{x\to0} \frac{2 (\sin x)^2}{x^2} = 2 * \lim_{x\to0} (\frac{\sin x}{x})^2 = 2 * 1 = 2$
Which way is true? Is it possible to help me?
I'm sorry for bad English.
Thanks. 

The second factor in $$\frac{\sin x-x\cos x}{x^3}\frac{\sin x+x\cos x}{x}$$ clearly tends to $2$.
Then by L'Hospital,
$$\frac{\cos x-\cos x+x\sin x}{3x^2}$$ tends to $1/3$.
