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

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.

• You can use Taylor series, see this post. Commented Nov 20, 2022 at 10:18
• Thank you! But why are the steps I did wrong? I believe I haven't done anything illegal Commented Nov 20, 2022 at 10:21
• You cannot add or substract equivalences $\sim$: for example, $x^{10}\sim x^{10}+1$ and $-x^{10}+1 \sim -x^{10}+2$ but $-1\not\sim 3$ Commented Nov 20, 2022 at 10:24
• @Taladris Thank you, I think I see what I did wrong now Commented Nov 20, 2022 at 10:28
• If $f_1 \sim g_1$ and $f_2 \sim g_2$, that does not mean that $f_1 + f_2 \sim g_1 + g_2$. Likewise for $f_1 - f_2$. Commented Nov 20, 2022 at 10:31

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