# How do you find the limit of inverse trig functions?

For example, $x = {f}^{- 1} \left(y\right) = {\sin}^{- 1} \left(y\right)$ is defined to be the inverse function of $y = f \left(x\right) = \sin \left(x\right)$ for $- \frac{\pi}{2} \setminus \le q x \setminus \le q \frac{\pi}{2}$. Since $y = f \left(x\right) = \sin \left(x\right)$ is continuous and $y \to 1$ as $x \to \setminus {\frac{\pi}{2}}^{-}$ (the minus sign to the right of the number indicates the number is being approached "from the left"), it follows that $x = {f}^{- 1} \left(y\right) = {\sin}^{- 1} \left(y\right) \to \frac{\pi}{2}$ as $y \to {1}^{-}$. Similarly, $x = {f}^{- 1} \left(y\right) = {\sin}^{- 1} \left(y\right) \to - \setminus \frac{\pi}{2}$ as $y \to - {1}^{+}$.
Since $x = {g}^{- 1} \left(y\right) = {\cos}^{- 1} \left(y\right)$ is defined to be the inverse function of $y = g \left(x\right) = \cos \left(x\right)$ for $0 \setminus \le q x \setminus \le q \pi$ and both functions are continuous, it follows that $x = {g}^{- 1} \left(y\right) = {\cos}^{- 1} \left(y\right) \to 0$ as $y \to {1}^{-}$ and $x = {g}^{- 1} \left(y\right) = {\cos}^{- 1} \left(y\right) \to \pi$ as $y \to - {1}^{+}$.
Since $x = {h}^{- 1} \left(y\right) = {\tan}^{- 1} \left(y\right)$ is defined to be the inverse function of $y = h \left(x\right) = \tan \left(x\right)$ for $- \frac{\pi}{2} < x < \frac{\pi}{2}$ and both functions are continuous, it follows that $x = {h}^{- 1} \left(y\right) = {\tan}^{- 1} \left(y\right) \to \frac{\pi}{2}$ as $y \to + \setminus \infty$ and $x = {h}^{- 1} \left(y\right) = {\tan}^{- 1} \left(y\right) \to - \frac{\pi}{2}$ as $y \to - \setminus \infty$.