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