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?
 A: $$\lim_{x\to 0} \frac{(4^{\arcsin(x^2)} - 1)(\sqrt[10]{1 - \arctan(3x^2)} - 1)}{(1-\cos\tan6x)\ln(1-\sqrt{\sin x^2})}
$$
$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 $0$.
A: 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
$$\lim_{x\to 0} \dfrac{x^2 \log (4) \cdot \left(-\frac{3 x^2}{10}\right)}{-18 x^2 \cdot x} = \lim_{x\to 0} \dfrac{3\log(4)}{180}x = 0$$
A: The limit of a product is the product of the limits (provided they exist). Multiply and divide by terms so that the resulting limmand is a product of known limits:
$$\frac{\left(4^{\arcsin(x^2)}-1\right)\left(\sqrt[10]{1-\arctan(3x^2)}-1\right)}{(1- \cos \tan 6x)\log(1-\sqrt{\sin x^2})}$$
$$= \frac{e^{\log 4 \arcsin x^2}-1}{\log 4\arcsin x^2}\cdot \log 4\cdot \frac{\arcsin x^2}{x^2}\cdot\frac{\left(\sqrt[10]{1-\arctan(3x^2)}\right)^{10}-1^{10}}{\arctan(3x^2)}\cdot\frac{1}{\left(\sqrt[10]{1-\arctan(3x^2)}\right)^{9}+\cdots+1^9}\cdot \frac{\arctan 3x^2}{3x^2} \cdot \frac{\tan^2 6x}{1-\cos \tan 6x}\cdot \frac{36x^2}{\tan^2 6x}\cdot \frac{\sqrt{\sin x^2}}{\log(1-\sqrt{\sin x^2})}\cdot\sqrt{\frac{x^2}{\sin x^2}}\cdot\frac{3}{36}\cdot|x|$$
$$
\longrightarrow 1 \cdot \log 4 \cdot 1 \cdot -1 \cdot \frac{1}{10} \cdot 1 \cdot 2 \cdot 1 \cdot 1 \cdot 1 \cdot \frac{1}{12} \cdot 0 = -\frac{\log 4}{60}\cdot 0$$
With an extra power of $|x|$ in the denominator, it converges to the values left over.
