I am confused about how we calculate this limit without L'Hopital rule.
$$\lim_{x\to0} \frac{\tan(x)-\sin(x)}{x^3}$$
The steps I was able to do are
$$
\lim_{x\to0} \frac{\tan(x)-\sin(x)}{x^3}=
\lim_{x\to0} \frac{\tan(x)}{x^3}-\frac{\sin(x)}{x^3}=
\lim_{x\to0} \frac{1}{x^2}-\frac{1}{x^2}=
\lim_{x\to0} 0 = 0
$$
However evaluating this limit using Wolfram Mathematica I get the result $\frac{1}{2}$.
I suspect the problem to be in the simplifications $\frac{\tan(x)}{x^3}\sim\frac{1}{x^2}$ and $\frac{\sin(x)}{x^3}\sim\frac{1}{x^2}$ but I don't understand how exactly.
Best Answer
Your suspection is right: you cannot use equivalence in sum terms, only in a multiplicative terms. In this particular case, given $\tan x = x +\color{green} {o(x)}$, $\sin x = x +\color{red} {o(x)}$, you have different $o(x)$ in them, so you cannot just cancel them out.
Way to solve this can be factor $\sin x$ out and using equivalence $\sin x\sim x$, have
$$ \lim_{x\to 0} \frac {\tan x - \sin x} {x^3} = \lim_{x\to 0} \frac {\frac 1 {\cos x}- 1}{x^2} = \lim_{x\to 0} \frac 1 {\cos x} \lim_{x\to 0} \frac {1-\cos x} {x^2} = \frac 1 2 $$