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. 
 A: Your first way is wrong, and the mistake is that you in fact argued that
$$\lim_{x\to0}\left(\frac{\sin x}x\right)^2=1\implies \lim_{x\to0}\sin^2x=x^2$$
and this is, of course, wrong.
A: I think this problem is very common among beginners dealing with limits. The meaning of $$\lim_{x \to 0}\frac{\sin x}{x} = 1\tag{1}$$ is not that you can replace $\sin x$ by $x$ as if $\sin x = x$ when $x \to 0$. The meaning of equation $(1)$ is that whenever you see the expression $\lim\limits_{x \to 0}\dfrac{\sin x}{x}$ then you can replace this entire expression by $1$. There is no more meaning to equation $(1)$ and any approach to give it more meaning will only lead to confusion (as evident from current question).
The right approach to evaluation of limits is the use of theorems dealing with limits. Such theorems include laws of algebra of limits (including the exact conditions under which they are applicable), limit of composition of functions (rule of substitution), Squeeze theorem, L'Hospital's Rule and Taylor's theorem with Peano's form of remainder.

The problem here can be easily solved by factorization of numerator and denominator as follows
\begin{align}
L &= \lim_{x \to 0}\frac{(\sin x)^{2} - (x\cos x)^{2}}{x^{4}}\notag\\
&= \lim_{x \to 0}\frac{\sin x - x\cos x}{x^{3}}\cdot\frac{\sin x + x \cos x}{x}\notag\\
&= \lim_{x \to 0}\frac{\sin x - x\cos x}{x^{3}}\cdot\lim_{x \to 0}\left(\frac{\sin x}{x} + \cos x\right)\notag\\
&= 2\lim_{x \to 0}\frac{\sin x - x\cos x}{x^{3}}\notag\\
&= 2\lim_{x \to 0}\frac{\sin x - x}{x^{3}} + \frac{1 - \cos x}{x^{2}}\notag\\
&= 2(A + B)\notag
\end{align}
The second limit $B$ in last step is easily calculated as $$B = \lim_{x \to 0}\frac{1 - \cos x}{x^{2}} = \lim_{x \to 0}\frac{1 - \cos^{2} x}{x^{2}(1 + \cos x)} = \frac{1}{2}\lim_{x \to 0}\frac{\sin^{2}x}{x^{2}} = \frac{1}{2}$$ The first limit $A$ can be calculated via Taylor's series (or via L'Hospital's Rule) as follows $$B = \lim_{x \to 0}\frac{\sin x - x}{x^{3}} = \lim_{x \to 0}\dfrac{x - \dfrac{x^{3}}{6} + o(x^{3}) - x}{x^{3}} = -\frac{1}{6}$$ and hence the desired limit $L$ is given by $L = 2(A + B) = 2/3$.

Someone may ask: the factorization in the numerator is obvious (based on $a^{2} - b^{2} = (a + b)(a - b)$) but why do we factor $x^{4}$ as $x^{3}\cdot x$ and perhaps not as $x^{2}\cdot x^{2}$? The answer is that one of the factors in numerator is $(\sin x + x\cos x)$ and this when divided by $x$ gives $((\sin x)/x + \cos x)$ which has a well defined limit $2$.
The next question is: why not pair the factor $x$ in denominator with factor $(\sin x - x\cos x)$ in numerator? After all this also leads to $((\sin x)/x - \cos x)$ which has a well defined limit $0$. This gives rise to an important fact about the laws of algebra of limits especially in case of product.
The usual statement of product law of limits says that if both $\lim_{x \to a}f(x)$ and $\lim_{x \to a}g(x)$ exist then $\lim_{x \to a}f(x)g(x)$ also exists and $$\lim_{x \to a}f(x)g(x) = \lim_{x \to a}f(x)\cdot\lim_{x \to a}g(x)\tag{2}$$ Thus the rule requires us to first verify that limits of both the factors $f(x)$ and $g(x)$ exist. A more convenient improvement on the above rule is the following:
If $\lim_{x \to a}f(x)$ exists and is non-zero then the following equation $$\lim_{x \to a}f(x)g(x) = \lim_{x \to a}f(x)\cdot\lim_{x \to a}g(x)\tag{3}$$ holds irrespective of the fact whether $\lim_{x \to a}g(x)$ exists or not. The meaning of equation $(3)$ is to be interpreted in this manner: if $\lim_{x \to a}g(x)$ exists then $\lim_{x \to a}f(x)g(x)$ also exists and its value is given by equation $(3)$. If $\lim_{x \to a}g(x)$ does not exist then the behavior of $\lim_{x \to a}f(x)g(x)$ is same as that of $\lim_{x \to a}g(x)$.
The above formulation of the product rule is more helpful in practice because it requires you to check the existence of limit of only one of the factors and if this limit is non-zero you can safely use the product rule. Hence when using the product rule it always makes sense to have one factor with non-zero limit especially when you are not sure about the limit of other factor. Note however that if the limit of a factor is $0$ then you must investigate/analyze the limit of other factor before reaching any conclusion about the limit of the product.
A: The correct way is the second one. The limit is indeed $\frac{2}{3}$.
Since$x$ is approaching to zero, you're lead to use Taylor expansion which is a cool tool to evaluate some limits. 
A: Note that we have

$$\begin{align}
\frac{\sin^2(x)-x^2\cos^2(x)}{x^4}&=\frac{\left(\frac{\sin(x)}{x}\right)^2-1}{x^2}+\frac{\sin^2(x)}{x^2}\\\\
&=\color{blue}{\underbrace{\left(\frac{\sin(x)}{x}+1\right)}_{\to 2\,\,\text{as}\,\,x\to 0}}\left(\frac{\sin(x)-x}{x^3}\right)+\color{red}{\underbrace{\frac{\sin^2(x)}{x^2}}_{\to 1\,\,\text{as}\,\,x\to 0}}\\\\
\end{align}$$

Evaluating the limit of interest boils down to evaluating the limit of $\frac{\sin(x)-x}{x^3}$.  
Given that $\sin(x)-x=-\frac16 x^3+O(x^5)$, the limit is easily seen to be $-\frac16$.  
Putting it all together, we find the limit of interest is

$$\lim_{x\to 0}\left(\frac{\sin^2(x)-x^2\cos^2(x)}{x^4}\right)=\frac23$$

A: If you are familiar with the Maclaurin series for the basic trig functions, you can solve this limit and others like it very quickly or even in your head. It also avoids differentiation mistakes from L'Hospital's. 
In your case, second order expansion is enough. Taking 
$$
\sin(x)\sim x-\frac{x^3}{6}\\
x\cos(x)\sim x(1-\frac{x^2}{2})=x-\frac{x^3}{2}
$$
Your limit becomes
$$
\lim_{x\to 0} \frac{(\sin x)^2 - (x \cos x) ^ 2}{x^4}=
\lim_{x\to 0} \frac{(x- \frac{x^3}{6})^2- (x-\frac{x^3}{2}) ^ 2}{x^4}
$$
And now you can notice that lower order terms terms dominate, and fortunately the $x^2$ cancel (or it would diverge), and pluck out the coefficients on $x^4$, which is $2/3$. If this is uncomfortable, you can always expand and check.
A: 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}
A: 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 $1/3$.
