Error calculating the limit $ \lim_{x \to 0}{\frac{x-\tan(x)}{x^2 \cdot \sin(x)}} $ Given this limit:  
$$ \lim_{x \to 0}{\frac{x-\tan(x)}{x^2 \cdot \sin(x)}} $$
Wolfram says the result is $\frac{1}{3}$ , but I tried to solve it and I get 0:  
$$ \lim_{x \to 0}{\frac{x \cdot (1-\frac{\tan(x)}{x})}{x \cdot (x \cdot \sin(x))}} = $$
$$ \lim_{x \to 0}{\frac{1-\frac{\tan(x)}{x}}{x \cdot \sin(x)}} = $$
$$ \lim_{x \to 0}{\frac{1}{x \cdot \sin(x)}} - \frac{1}{x \cdot \sin(x)} = 0$$
This because $ \lim_{x \to 0}{\frac{\tan(x)}{x}} = 1$
But the result is wrong.  
I would like to know not only it's right solution, but also (and specially) what's wrong in this attempt to solve it.
 A: By the Taylor series we have
$$\tan x=x+\frac{x^3}{3}+O\left(x^5\right)$$
and
$$\sin x=x+O(x^2)$$
so
$$\lim_{x\to0}\frac{x-\tan(x)}{x^2 \cdot \sin(x)}=\lim_{x\to0}\frac{-\frac{x^3}{3}+O\left(x^5\right)}{x^3+O(x^4)}=-\frac{1}{3}$$
A: Sombody else has already shown you a way to compute the limit yourself; let me explain the problem with your attempt.
The issue here is that this last step is in an indeterminate form "$\infty-\infty$", which need not tend to 0.  
Think about it this way: consider $x^2$ and $(1+\frac{1}{x})x^2$. As $x\rightarrow\infty$, clearly $1+\frac{1}{x}\rightarrow1$; but
$$
\left(1+\frac{1}{x}\right)x^2-x^2=x\rightarrow\infty\text{ as }x\rightarrow\infty.
$$
A: First, let's analyze your attempt.
This example has the same problem:
$$
\begin{align}
\lim_{x\to0}\frac{(1+x)-1}{x}
&=\lim_{x\to0}\left(\frac{1+x}{x}-\frac{1}{x}\right)\\
&=\lim_{x\to0}\left(\frac{1}{x}-\frac{1}{x}\right)\\[6pt]
&=0
\end{align}
$$
It is not permitted to take the limit of a piece of an expression before the other parts.
On the other hand, since we are subtracting exact equals, we can do
$$
\begin{align}
\lim_{x\to0}\frac{(1+x)-1}{x}
&=\lim_{x\to0}\left(\frac{1+x}{x}-\frac{1}{x}\right)\\
&=\lim_{x\to0}\left(\frac{x}{x}\right)\\[6pt]
&=1
\end{align}
$$

Now, let's evaluate the limit is a couple of ways.
Without using calculus, we cam work from this answer and subtract $(10)$ from $(9)$ to get
$$
\lim_{x\to0}\frac{\tan(x)-x}{x^3}=\frac13
$$
Divide by
$$
\lim_{x\to0}\frac{\sin(x)}{x}=1
$$
(proven geometrically here) to get
$$
\lim_{x\to0}\frac{\tan(x)-x}{x^2\sin(x)}=\frac13
$$
We can also use L'Hospital
$$
\begin{align}
\lim_{x\to0}\frac{\tan(x)-x}{x^2\sin(x)}
&=\lim_{x\to0}\frac{\tan(x)-x}{x^3}\lim_{x\to0}\frac{x}{\sin(x)}\\
&=\lim_{x\to0}\frac{\sec^2(x)-1}{3x^2}\lim_{x\to0}\frac{1}{\cos(x)}\\
&=\lim_{x\to0}\frac{2\sec^2(x)\tan(x)}{6x}\cdot1\\
&=\frac13\lim_{x\to0}\sec^2(x)\lim_{x\to0}\frac{\tan(x)}{x}\\
&=\frac13\cdot1\cdot\lim_{x\to0}\frac{\sec^2(x)}{1}\\
&=\frac13
\end{align}
$$
A: As  $$ \lim_{x \to 0}{\frac{x-\tan(x)}{x^2 \cdot \sin(x)}} $$ is of the form $\frac\infty\infty,$
We can utilize L'Hôpital's rule, too.
So, $$ \lim_{x \to 0}{\frac{x-\tan(x)}{x^2 \cdot \sin(x)}}=\lim_{x \to 0}\frac{1-\sec^2x}{2x\sin x+x^2\cos x}=-\lim_{x \to 0}\frac{\tan^2x}{2x\sin x+x^2\cos x}$$ 
$$=-\lim_{x \to 0}\frac{\sin^2x}{\cos^2x(2x\sin x+x^2\cos x)}$$ 
$$= -\left(\lim_{x \to 0}\frac{\sin x}x\right)^2\cdot\frac1{\lim_{x \to 0}\cos^2x\cdot \left(2\lim_{x \to 0}\frac{\sin x}x+\lim_{x \to 0}\cos x\right)}$$ (Dividing the numerator & the denominator by $x^2$ )
Now, we know, $\lim_{h\to0}\frac{\sin h}h=?$ and $\lim_{h\to0}\cos h=?$
A: $$
\begin{align}
&\lim_{x \to 0}{\frac{1-\frac{\tan(x)}{x}}{x \cdot \sin(x)}}\\
&=\lim_{x \to 0}{\left(\frac{1}{x \cdot \sin(x)}-\frac{\frac{\tan(x)}{x}}{x \cdot \sin(x)}\right)}\\
&\neq\lim_{x \to 0}{\left(\frac{1}{x \cdot \sin(x)}-\frac{\lim_{x \to 0}{\frac{\tan(x)}{x}}}{x \cdot \sin(x)}\right)}\\
&=\lim_{x \to 0}{\left(\frac{1}{x \cdot \sin(x)}-\frac{1}{x \cdot \sin(x)}\right)}\\
&=\lim_{x \to 0}{0}=0
\end{align}
$$
