Find limit of trigonometric function with indeterminacy Find limit of the given function:
$$\lim_{x\rightarrow0} \frac{(4^{\arcsin(x^2)} - 1)(\sqrt[10]{1 - \arctan(3x^2)} - 1)}{(1-\cos\tan6x)\ln(1-\sqrt{\sin x^2})}
$$
I tried putting 0 instead of x
$$\lim_{x\rightarrow0} \frac{(4^{\arcsin(x^2)} - 1)(\sqrt[10]{1 - \arctan(3x^2)} - 1)}{(1-\cos\tan6x)\ln(1-\sqrt{\sin x^2})}
=
\frac{(4^{\arcsin(0)} - 1)(\sqrt[10]{1 - \arctan(0)} - 1)}{(1-\cos\tan0)\ln(1-\sqrt{\sin 0})}
=
\frac{(1 - 1)(\sqrt[10]{1 - 0} - 1)}{(1-1)\ln(1-\sqrt{0})}
=
\frac{0*0}{0*0}
=
\frac{0}{0}
$$
But as you can see at $0$ there is limit Indeterminacy ($0/0$). How to play around and solve it?

$$\lim_{x\to 0} \frac{(4^{\arcsin(x^2)} - 1)(\sqrt[10]{1 - \arctan(3x^2)} - 1)}{(1-\cos\tan6x)\ln(1-\sqrt{\sin x^2})}
$$
$4^{\arcsin(x^2)} - 1=e^{\arcsin(x^2) \log 4}-1\sim x^2\log 4 $
$\sqrt[10]{1 - \arctan(3x^2)} - 1\sim 1-\dfrac{1}{10}\, \arctan(3x^2)-1\sim -\dfrac{3}{10}x^2 $
$1-\cos\tan6x \sim 1-1+\dfrac{\tan^2 6x}{2}\sim 18 x^2$
$\ln(1-\sqrt{\sin x^2})\sim \sqrt{\sin x^2} \sim |x|$
Therefore the limit is $0$.
