<s>[INST] 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
 [/INST] I think the limit notation has been made so confusing thanks to various educators (including book authors) who try to simplify it too much. When you say $$\lim_{x \to 0}\frac{\sin x}{x} = 1 \tag{1}$$ you don't mean $(\sin x)/x = 1$ when $x \to 0$. Hence it is just not possible to replace $\sin x$ by $x$. However whenever you see the expression $\lim_{x \to 0}\dfrac{\sin x}{x}$ you can replace it by $1$ (because of the equation $(1)$ above).
You can always replace a thing called $A$ by a thing called $B$ when you know that $A$ and $B$ are equal but if $A \neq B$ then you can't replace $A$ by $B$. Unfortunately people are so used to invalid replacements while calculating limits that they don't notice that it is wrong.
The way to solve your limit problem is as follows:
\begin{align}
L &= \lim_{x \to 0}\frac{x - \sin x}{\sin 2x - \tan 2x}\notag\\
&= \lim_{x \to 0}\dfrac{x - \sin x}{\sin 2x - \dfrac{\sin 2x}{\cos 2x}}\notag\\
&= \lim_{x \to 0}\dfrac{x - \sin x}{\sin 2x}\cdot\frac{\cos 2x}{\cos 2x - 1}\notag\\
&= \lim_{x \to 0}\dfrac{x - \sin x}{\dfrac{\sin 2x}{2x}\cdot 2x}\cdot\frac{\cos 2x}{\cos 2x - 1}\notag\\
&= \lim_{x \to 0}\dfrac{x - \sin x}{2x(\cos 2x - 1)}\cdot\dfrac{1}{\lim_\limits{x \to 0}\dfrac{\sin 2x}{2x}}\cdot \lim_{x \to 0}\cos 2x\notag\\
&= \frac{1}{2}\lim_{x \to 0}\frac{x - \sin x}{x(\cos 2x - 1)}\cdot 1\cdot 1\notag\\
&= \frac{1}{2}\lim_{x \to 0}\frac{\sin x - x}{2x\sin^{2}x}\notag\\
&= \frac{1}{4}\lim_{x \to 0}\frac{\sin x - x}{x^{3}}\cdot\frac{x^{2}}{\sin^{2}x}\notag\\
&= \frac{1}{4}\lim_{x \to 0}\frac{\sin x - x}{x^{3}}\cdot\lim_{x \to 0}\frac{x}{\sin x}\cdot\lim_{x \to 0}\frac{x}{\sin x}\notag\\
&= \frac{1}{4}\lim_{x \to 0}\frac{\sin x - x}{x^{3}}\cdot 1\cdot 1\notag\\
&= \frac{1}{4}\lim_{x \to 0}\frac{\sin x - x}{x^{3}}\notag\\
&= \frac{1}{4}\lim_{x \to 0}\frac{\cos x - 1}{3x^{2}}\text{ (via L'Hospital's Rule)}\notag\\
&= \frac{1}{12}\lim_{x \to 0}\frac{\cos^{2}x - 1}{x^{2}(1 + \cos x)}\notag\\
&= -\frac{1}{12}\lim_{x \to 0}\frac{\sin^{2}x}{x^{2}}\cdot\lim_{x \to 0}\frac{1}{1 + \cos x}\notag\\
&= -\frac{1}{12}\cdot 1\cdot\frac{1}{2}\notag\\
&= -\frac{1}{24}\notag
\end{align}
The above solution has been presented in a slightly more detailed fashion (than usually necessary for an exam) in order to show how the expression $\lim_{x \to 0}\dfrac{\sin x}{x}$ is replaced by $1$ (and how $\lim_{x \to 0}\cos x$ is replaced by $1$).
</s>