Not getting the same solution when using the rule sin(x)\x=1 on a limit There is a rule in limits that when $x$ approaches zero:
$$\frac{\sin\left(x\right)}{x}=1$$
So I used this rule on the following exercise:

Evaluate
  $$
\lim _{x\to 0}\:\frac{x-\sin\left(x\right)}{\sin\left(2x\right)-\tan\left(2x\right)}
$$    

I substituted $\sin(2x)$ with $2x$ by the following way:
$$\sin\left(2x\right)=\frac{\sin\left(2x\right)}{2x}\cdot 2x=1\cdot 2x=2x \Rightarrow \lim _{x\to 0}\:\frac{x-\sin\left(x\right)}{2x-\tan\left(2x\right)}$$
But according to symbolab:
$$\lim _{x\to 0}\:\frac{x-\sin\left(x\right)}{2x-\tan\left(2x\right)}=-\frac{1}{16}$$
while
$$\lim _{x\to 0}\:\frac{x-\sin\left(x\right)}{\sin(2x)-\tan\left(2x\right)}=-\frac{1}{24}$$

Why am I getting this contradiction?
More over if I susbsitute the following I do get the right answer
$$\tan\left(2x\right)=\frac{\sin\left(2x\right)}{\cos\left(2x\right)}=\frac{2x}{\cos\left(2x\right)}\Rightarrow \lim \:_{x\to \:0}\:\frac{x-\sin\left(x\right)}{2x-\frac{2x}{\cos\left(2x\right)}}=-\frac{1}{24}$$
If you want to test yourself: Symbolab with the excersice preloaded

The problem you encountered was that both $\sin(x)-\tan(x)=O\!\left(x^3\right)$ and $x-\sin(x)=O\!\left(x^3\right)$, this means that substituting $x$ for $\sin(x)$ may substantively change the limit.
To be precise,
$$
\begin{align}
\sin(x)-\tan(x)
&=\left(x-\frac{x^3}6+O\!\left(x^5\right)\right)-\left(x+\frac{x^3}3+O\!\left(x^5\right)\right)\\
&=-\frac{x^3}2+O\!\left(x^5\right)
\end{align}
$$
whereas
$$
\begin{align}
x-\tan(x)
&=x-\left(x+\frac{x^3}3+O\!\left(x^5\right)\right)\\
&=-\frac{x^3}3+O\!\left(x^5\right)
\end{align}
$$
This explains why the limits you got differ by a factor of $\frac32$.
