Simple limit. Used L'Hôpital's rule. Didn't work.

Question

$\lim_{x\rightarrow 0}f(x)$

$f(x)=\frac{\exp (\arcsin \left (x \right ))-\exp (\sin \left (x \right ))}{\exp (\arctan \left (x \right ))-\exp (\tan \left (x \right ))}$

I tried the L'Hôpital's rule as mentioned in the title and replaced ($\arcsin x$,$\arctan x$,$\sin x$,$\tan x$) with $\sim_{0}$ $x$ but still have an indeterminate form. Any hints please ?

Answer 1

Alternatively, one may use Taylor series expansions, as $x \to 0$, $$ \begin{align} \sin x=& x-\frac{x^3}6+o(x^3) &\tan x=& x+\frac{x^3}3+o(x^3) \\\arcsin x=& x+\frac{x^3}6+o(x^3) &\arctan x=& x-\frac{x^3}3+o(x^3) \end{align} $$ and $$ e^u=1+u+\frac{u^2}2+\frac{u^3}6+o(u^3),\quad u \to0, $$giving, as $x \to 0$, $$ \frac{e^{\arcsin x}-e^{\sin x}}{e^{\arctan x}-e^{\tan x}}=\frac{\frac{x^3}3+o(x^3)}{-\frac{2x^3}3+o(x^3)}=\color{blue}{-\frac12}+o(1). $$

Answer 2

We can proceed as follows \begin{align} L &= \lim_{x \to 0}\frac{e^{\arcsin x} - e^{\sin x}}{e^{\arctan x} - e^{\tan x}}\notag\\ &= \lim_{x \to 0}e^{\sin x - \tan x}\cdot\frac{e^{\arcsin x - \sin x} - 1}{e^{\arctan x - \tan x} - 1}\notag\\ &= \lim_{x \to 0}\frac{e^{\arcsin x - \sin x} - 1}{\arcsin x - \sin x}\cdot\frac{\arcsin x - \sin x}{\arctan x - \tan x}\cdot\frac{\arctan x - \tan x}{e^{\arctan x - \tan x} - 1}\notag\\ &= \lim_{x \to 0}\frac{\arcsin x - \sin x}{\arctan x - \tan x}\notag\\ &= \lim_{x \to 0}\dfrac{\left(x + \dfrac{x^{3}}{6} + o(x^{3})\right) - \left(x - \dfrac{x^{3}}{6} + o(x^{3})\right)}{\left(x - \dfrac{x^{3}}{3} + o(x^{3})\right) - \left(x + \dfrac{x^{3}}{3} + o(x^{3})\right)}\notag\\ &= -\frac{1}{2}\notag \end{align}