# How do you find the limit of arcsin(x)/ arctan(x) as x approaches 0?

${\lim}_{x \rightarrow 0} \frac{{\sin}^{-} 1 x}{{\tan}^{-} 1 x} = 1$

#### Explanation:

We evaluate the limit first

${\lim}_{x \rightarrow 0} \frac{{\sin}^{-} 1 x}{{\tan}^{-} 1 x} = \frac{0}{0}$

We apply L'Hospitals Rule

${\lim}_{x \rightarrow 0} \frac{{\sin}^{-} 1 x}{{\tan}^{-} 1 x} = {\lim}_{x \rightarrow 0} \frac{\frac{d}{\mathrm{dx}} \left({\sin}^{-} 1 x\right)}{\frac{d}{\mathrm{dx}} \left({\tan}^{-} 1 x\right)} = {\lim}_{x \rightarrow 0} \frac{\left(\frac{1}{\sqrt{1 - {x}^{2}}}\right)}{\left(\frac{1}{1 + {x}^{2}}\right)} = 1$

God bless....I hope the explanation is useful.