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. 

Your first way is wrong and, unfortunately, a common mistake. With the same argument you would conclude that
$$
\lim_{x\to0}\frac{x-\sin x}{x^3}=0
$$
which is a big error (see below).
You can change $\sin x$ into $x$ when it's a factor, not a summand. For instance, if you have
$$
\lim_{x\to0}\frac{\sqrt{1-x}-1}{\sin x}
$$
you can as well compute
$$
\lim_{x\to0}\frac{\sqrt{1-x}-1}{x}=
\lim_{x\to0}\frac{1-x-1}{x(\sqrt{1-x}+1)}=-\frac{1}{2}
$$
because then
$$
\lim_{x\to0}\frac{\sqrt{1-x}-1}{\sin x}=
\lim_{x\to0}\frac{\sqrt{1-x}-1}{x}\frac{x}{\sin x}=
\lim_{x\to0}\frac{\sqrt{1-x}-1}{x}\lim_{x\to0}\frac{x}{\sin x}=-\frac{1}{2}\cdot 1
$$
For the second way you have used Taylor expansions and, indeed, the first limit above can be computed correctly with them:
$$
\lim_{x\to0}\frac{x-\sin x}{x^3}=
\lim_{x\to0}\frac{x-\left(x-\dfrac{x^3}{6}+o(x^3)\right)}{x^3}=\frac{1}{6}
$$
Do you see? The quotient $\dfrac{\sin x}{x}$ is “like $1$”, but the difference $x-\sin x$ is “like $\dfrac{x^3}{6}$”.
You can tackle the computation of the limit by observing that
\begin{align}
(\sin x)^2-(x\cos x)^2
&=(\sin x-x\cos x)(\sin x+x\cos x) \\
&=\left(x-\frac{x^3}{6}+o(x^3)-x\left(1-\frac{x^2}{2}+o(x^2)\right)\right)
  (x+o(x)+x(1+o(1)) \\
&=\left(\frac{x^3}{3}+o(x^3)\right)(2x+o(x)) \\
&=\frac{2}{3}x^4+o(x^4)
\end{align}
