Why is this limit $\frac{1}{2}$?

Question

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.

Answer 1

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

$$ \lim_{x\to 0} \frac {\tan x - \sin x} {x^3} = \lim_{x\to 0} \frac {\frac 1 {\cos x}- 1}{x^2} = \lim_{x\to 0} \frac 1 {\cos x} \lim_{x\to 0} \frac {1-\cos x} {x^2} = \frac 1 2 $$

Answer 2

The problem is you are bypassing an indeterminate form; in fact: $$\lim_{x\to 0} \dfrac{\tan x-\sin x}{x^3}= \lim_{x\to 0} \dfrac{\tan x}{x^3}-\dfrac{\sin x}{x^3}=\lim_{x\to 0}\dfrac{1}{x^2}\left(\dfrac{\tan x}{x}-\dfrac{\sin x}{x} \right)$$ Now $x^2 \to 0 \,$ and so does $\left(\dfrac{\tan x}{x}-\dfrac{\sin x}{x} \right)$; this is the problem.
Instead, using Taylor expansion and knowing that $\tan x \sim x+\dfrac{x^3}{3}$ and $\sin x \sim x-\dfrac{x^3}{6}$ you have: $\dfrac{\tan x-\sin x}{x^3} \sim \dfrac{x^3}{2x^3}\to \dfrac{1}{2}\quad$ as $x \to 0$