Limit involving exponentials of $\arcsin(x)$ and $\arctan(x)$

Question

How can I calculate this limit without L'Hospital rule and Taylor series?

$$ \lim_{x \to 0} \frac{e^{\arctan(x)} - e^{\arcsin(x)}}{1 - \cos^3(x)} $$

Answer 1

Let $L=\lim_{x \to 0} \frac{e^{\arctan(x)} - e^{\arcsin(x)}}{1 - \cos^3(x)}$
$L=\lim_{x \to 0} \frac{(e^{\arctan(x)} -1)- (e^{\arcsin(x)}-1)}{1 - \cos^3(x)}$
$L=\lim_{x \to 0} \frac{\arctan (x)\frac{(e^{\arctan(x)} -1)}{\arctan(x)}- \arcsin (x)\frac{(e^{\arcsin(x)} -1)}{\arcsin(x)}}{1 - \cos^3(x)}$
$L=\lim_{x \to 0}\frac{\arctan (x)-\arcsin (x)}{1 - \cos^3(x)}$
$L=\lim_{x \to 0}\frac{\arctan (x)-\arcsin (x)}{(1 - \cos(x))(\cos^2x+\cos x+1)}$

$L=\lim_{x \to 0}\frac{\arctan (x)-\arcsin(x)}{x^2\frac{1-\cos x}{x^2}(\cos^2x+\cos x+1)}$
Now use $\lim_{x \to 0}\frac{1-\cos x}{x^2}=\frac{1}{2}$
$L=\lim_{x \to 0}\frac{\arctan (x)-\arcsin(x)}{x^2\times\frac{3}{2}}.....(1)$
Now let $L_1=\lim_{x \to 0}\frac{\arctan (x)-\arcsin(x)}{x^2}$
$L_1=\lim_{x \to 0}\frac{\arctan (2x)-\arcsin(2x)}{(2x)^2}$
$4L_1=\lim_{x \to 0}\frac{\arctan (2x)-\arcsin(2x)}{x^2}$
$4L_1-L_1=\lim_{x \to 0}\frac{\arctan (2x)-\arcsin(2x)}{x^2}-\lim_{x \to 0}\frac{\arctan (x)-\arcsin(x)}{x^2}$
$3L_1=\lim_{x \to 0}\frac{(\arctan 2x-\arctan x)-(\arcsin 2x-\arcsin x)}{x^2}$
$3L_1=\lim_{x \to 0}\frac{(\arctan \frac{x}{1+x^2})-\arcsin(2x\sqrt{1-x^2}-x\sqrt{1-4x^2})}{x^2}$
$3L_1=\lim_{x \to 0}\frac{\frac{x}{1+x^2}\frac{\arctan \frac{x}{1+x^2}}{\frac{x}{1+x^2}}-(2x\sqrt{1-x^2}-x\sqrt{1-4x^2})\frac{\arcsin(2x\sqrt{1-x^2}-x\sqrt{1-4x^2})}{2x\sqrt{1-x^2}-x\sqrt{1-4x^2}}}{x^2}$
$\lim_{x \to 0}\frac{\arctan \frac{x}{1+x^2}}{\frac{x}{1+x^2}}=1$ and

$\lim_{x \to 0}\frac{\arcsin(2x\sqrt{1-x^2}-x\sqrt{1-4x^2})}{(2x\sqrt{1-x^2}-x\sqrt{1-4x^2})}=1$
$3L_1=\lim_{x \to 0}\frac{\frac{x}{1+x^2}-(2x\sqrt{1-x^2}-x\sqrt{1-4x^2})}{x^2}$
$3L_1=\lim_{x \to 0}\frac{x}{(1+x^2)x^2}-\frac{(2\sqrt{1-x^2}-\sqrt{1-4x^2})}{x}$
$3L_1=\lim_{x \to 0}\frac{1}{(1+x^2)x}-\frac{({4-4x^2}-1+4x^2)}{x(2x\sqrt{1-x^2}+x\sqrt{1-4x^2})}$
$3L_1=\lim_{x \to 0}\frac{1}{(1+x^2)x}-\frac{1}{x}$
$3L_1=\lim_{x \to 0}\frac{1-1-x^2}{(1+x^2)x}$
$3L_1=\lim_{x \to 0}\frac{-x}{(1+x^2)}=0$
$L_1=0$
$L=\frac{2L_1}{3}=0$