Evaluation Of Trigonometric Limits (Form-3)

Generally to evaluate limits involving inverse trigonometric functions we convert the limit in terms of trigonometric functions by replacing the inverse trigonometric function by θ

$\displaystyle \left( i \right) \lim _{ x\longrightarrow 0 }{ \frac { \sin ^{ -1 }{ x } }{ x } } =1\\ Let\quad \sin ^{ -1 }{ x } =\theta .\\ Then\quad x=\sin { \theta } \\ Also\quad x\longrightarrow 0\Rightarrow \sin { \theta } \longrightarrow 0\Rightarrow \theta \longrightarrow 0\\ \therefore \lim _{ x\longrightarrow 0 }{ \frac { \sin ^{ -1 }{ x } }{ x } } \\ =\lim _{ x\longrightarrow 0 }{ \frac { \theta }{ \sin { \theta } } } \\ =\lim _{ x\longrightarrow 0 }{ \frac { 1 }{ \frac { \sin { \theta } }{ \theta } } } \\ =\frac { 1 }{ 1 } \\ =1$

$\displaystyle \left( ii \right) \lim _{ x\longrightarrow 0 }{ \frac { \tan ^{ -1 }{ x } }{ x } } =1\\ Let\quad \tan ^{ -1 }{ x } =\theta .\\ Then\quad x=\tan { \theta } \\ Also\quad x\longrightarrow 0\Rightarrow \tan { \theta } \longrightarrow 0\Rightarrow \theta \longrightarrow 0\\ \therefore \lim _{ x\longrightarrow 0 }{ \frac { \tan ^{ -1 }{ x } }{ x } } \\ =\lim _{ x\longrightarrow 0 }{ \frac { \theta }{ \tan { \theta } } } \\ =\lim _{ x\longrightarrow 0 }{ \frac { 1 }{ \frac { \tan { \theta } }{ \theta } } } \\ =\frac { 1 }{ 1 } \\ =1$

Example

$\displaystyle \frac { 1 }{ x } \sin ^{ -1 }{ \left( \frac { 2x }{ 1+{ x }^{ 2 } } \right) } \\ Let\quad x=\tan { \theta } .Then\quad x=0\Rightarrow \tan { \theta } =0\quad \Rightarrow \theta =0\\ \therefore \lim _{ x\longrightarrow 0 }{ \frac { 1 }{ x } \sin ^{ -1 }{ \left( \frac { 2x }{ 1+{ x }^{ 2 } } \right) } } \\ =\lim _{ \theta \longrightarrow 0 }{ \frac { 1 }{ \tan { \theta } } \sin ^{ -1 }{ \left( \frac { 2\tan { \theta } }{ 1+{ tan }^{ 2 }\theta } \right) } } \\ =\lim _{ \theta \longrightarrow 0 }{ \frac { 1 }{ \tan { \theta } } \sin ^{ -1 }{ \left( \sin { 2\theta } \right) } } \\ =\lim _{ \theta \longrightarrow 0 }{ \frac { 2\theta }{ \tan { \theta } } } \\ =2\left( 1 \right) \\ =2$