<s>[INST] $ \lim_{x\to 0}\frac{\tan x-\sin x}{\sin(x^3)}$ $$ \lim_{x\to 0}\frac{\tan x-\sin x}{\sin(x^3)} =[1]\lim_{x\to 0}\frac{\sin x/\cos x-\sin x}{x^3}\\ =[2]\lim_{x\to 0}\frac{1-\cos x}{x^2}\\ =\frac{1}{2} $$
My question is how [1]=[2]?
$\tan x-\sin x=\tan x(1-\cos x)=x(1-\cos x)$?
 [/INST] We have $\frac{\sin x}{\cos x}-\sin x=\frac{\sin x}{\cos x}(1-\cos x)$.
Thus our function is equal to
$$\frac{1}{\cos x}\frac{\sin x}{x}\frac{1-\cos x}{x^2}.$$
The limit of $\frac{1}{\cos x}$ is $1$. So is the limit of $\frac{\sin x}{x}$. 
You did not ask about the limit of $\frac{1-\cos x}{x^2}$. That can be obtained by multiplying top and bottom by $1+\cos x$, getting $\frac{1-\cos x}{x^2}=\frac{1}{1+\cos x}\left(\frac{\sin x}{x}\right)^2$. 
Remark: The above answered your specific question. It did not address the question of finding
$$\lim_{x\to 0} \frac{\tan x-\sin x}{\sin(x^3)}.\tag{1}$$
For that, we rewrite (1) as
$$\lim_{x\to 0} \frac{\tan x-\sin x}{x^3}\frac{x^3}{\sin(x^3)}.$$
Since $\lim_{t\to 0}\frac{\sin t}{t}=1$, we have $\lim_{x\to 0}\frac{x^3}{\sin(x^3)}=1$, and we are at the question you asked. 
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