limit of sin^-1(x)/x as x approaches to 0

Question : limit of sin^-1(x)/x as x approaches to 0 Solution : We shall let the limit is \[L={\rm limit}_{x\rightarrow0}{\frac{\sin^{-1}{\left(x\right)}}{x}}\] Since \[\sin^{-1}{\left(0\right)}=0\ Hence \frac{\sin^{-1}{\left(0\right)}}{0}=\frac{0}{0}\] We can apply L’Hospital rule, that is \[L={\rm limit}_{x\rightarrow0}{\frac{\frac{d}{dx}\sin^{-1}{\left(x\right)}}{\frac{d}{dx}x}} ={\rm limit}_{x\rightarrow0}{\frac{\frac{1}{\sqrt{1-x^2}}}{1}}\] \[=\frac{1}{\frac{\sqrt{1-0^2}}{1}}=1\] Therefore \[{\rm limit}_{x\rightarrow0}{\frac{\sin^{-1}{\left(x\right)}}{x}}=1\]

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