Trigonometric limit mistake Question: $$\lim_{x\to 0}\frac{\tan x-\sin x}{x^3}$$
The answer, by L'Hopital's rule as well as wolfram and desmos is $\frac{1}{2}$
Here's what I did:
$$\lim_{x\to0}({\tan x \over x}\times{1\over x^2}-{\sin x \over x}\times{1\over x^2})$$
$$\lim_{x\to0}({1 \over x^2}-{1 \over x^2})=0$$
Im not sure where the mistake is.

Note that there is no split of the limit like $$\lim f(x) - \lim g(x) $$ here. Rather the expression under limit has been split as a difference based on laws of algebra and this is perfect.
The mistake happens in the next step and is very common and that is replacing the expressions $(\sin x) /x$ and $(\tan x) /x$ by $1$. And that's just plain wrong. We all know that these expressions are never equal to $1$ and thus they can't be replaced by $1$. I really find it surprising that the mistake is so common inspite of the very obvious mathematical fact that one can't replace $A$ by $B$ unless $A=B$.
Well what you can really do is that you can always replace the expression $\lim_{x\to 0}\dfrac{\sin x} {x}$ with $1$ without any restrictions simply because they are equal. This emphasizes the fact the expression $\dfrac{\sin x} {x} $ is different from the expression $\lim_{x\to 0}\dfrac{\sin x} {x} $. Unless this simple fact is taken into consideration one can get into trouble.
I have described this problem in detail in this answer which also describes when such replacements are valid. 
