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 first order terms cancel in nominator and denominator, so the contributions for the limit are in the third order terms. Replace $\sin(x)$ with $x-\frac16x^3+x^5r(x)$ to get the correct result.



Transforming the term gives
$$
\frac{\cos 2x}{\cos 2x-1}·\frac{x-\sin x}{2\sin x\cos x}
=
\frac{\cos 2x}{\cos x}·\frac{x-\sin x}{-2\sin^2 x·2\sin x}
$$
The limit of the cosines is $1$, in the denominator one can replace $\sin x$ with $x$ and in the numerator the third order formula to get
$$
\frac{x-(x-\frac16 x^3+O(x^5))}{-4x^3+O(x^5)} = -\frac{1-O(x^2)}{24}
$$
When taking limits of an expression you cannot arbritraily replace parts of the expression in isolation. You need to calculate the limit of the entire expression.

In your case you could use l'Hopital three times to get:
$$\begin{align}
\lim_{x\to0}\:\frac{x-\sin(x)}{\sin(2x)-\tan(2x)}&amp;=\lim_{x\to0}\:\frac{1-\cos(x)}{2\cos(2x)-2\sec^2(2x)}\\
&amp;=\lim_{x\to0}\:\frac{\sin(x)}{-4\sin(2x)-8\tan(2x)\sec^2(2x)}\\
&amp;=\lim_{x\to0}\:\frac{\cos(x)}{-8\cos(2x)-16\sec^4(2x)-32\tan^2(2x)\sec^2(2x)}
\end{align}$$
Now you can simply replace $x$ with $0$ to obtain:
$$
\lim_{x\to0}\:\frac{\cos(x)}{-8\cos(2x)-16\sec^4(2x)-32\tan^2(2x)\sec^2(2x)}=\frac{1}{-8*1-16*1-32*0*1}=-\frac{1}{24}
$$