I am taking a Calculus 1 class and one exercise asks to solve the following limit:
$$ \lim_{x \to 0} \frac{\sin(\tan x) - \tan(\sin x)}{\arcsin(\arctan x) - \arctan(\arcsin x)} $$
Seeing as both parts of the division go to $0$, I decided to try L'Hopital, but after deriving everything twice (and still getting $\frac{0}{0}$) I gave up (it didn't seem like it would get me anywhere).
After that I tried the following:
$$ g(x) = \arcsin(\arctan x) - \arctan(\arcsin x) $$ $$ \lim_{x \to 0} \frac{\sin(\tan x)}{g(x)} - \frac{\tan (\sin x)}{g(x)} $$ $$ \lim_{x \to 0} \frac{\sin(\tan x)}{\tan x} \frac{\tan x}{g(x)} - \frac{\tan (\sin x)}{\sin x} \frac{\sin x}{g(x)} $$
And since $\lim_{x \to 0} \frac{\sin(\tan x)}{\tan x} = 1$ and $\lim_{x \to 0} \frac{\tan (\sin x)}{\sin x} = 1$:
$$ \lim_{x \to 0} \frac{\tan x}{g(x)} - \frac{\sin x}{g(x)} $$ $$ \lim_{x \to 0} \frac{\tan x}{x} \frac{x}{g(x)} - \frac{\sin x}{x} \frac{x}{g(x)} $$ $$ \lim_{x \to 0} \frac{x}{g(x)} - \frac{x}{g(x)} = \lim_{x \to 0} 0 = 0 $$
But the bad part is that this is wrong, the answer should be 1 and not 0. I have no clue where I went wrong, and worse still, I have no clue how to solve this exercise another way. Where did I go wrong? How could I solve this?
$$\lim_{x\to0}\left[\frac{\sin(\operatorname{tg}(x))}{\operatorname{tg}(x)}\cdot\frac{\operatorname{tg}(x)}x-\frac{\operatorname{tg}(\sin(x))}{\sin(x)}\cdot\frac{\sin(x)}x\right]\frac{x}{g(x)}$$
The difference tends to $0$, but $\frac x{g(x)}$ tends to $-\infty$ when $x\to 0$.
– peterwhy Jun 28 '23 at 03:42