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?

As the comment suggests, you shall use Taylor expansion up to $O(3)$, that is:
$$4^{\arcsin(x^2)} \approx 1+x^2 \log (4)+O\left(x^4\right)$$
$$\sqrt[10]{1- \arctan(3x^2)} \approx 1-\frac{3 x^2}{10}+O\left(x^4\right)$$
$$\cos(\tan(6x)) \approx 1-18 x^2+O\left(x^4\right)$$
$$\sqrt{\sin(x^2)} \approx x + O(x^4)$$
Plug the values inside and you will get
$$\lim_{x\to 0} \dfrac{x^2 \log (4) \cdot \left(-\frac{3 x^2}{10}\right)}{-18 x^2 \cdot x} = \lim_{x\to 0} \dfrac{3\log(4)}{180}x = 0$$
