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

May 19, 2015

It's based on how they are defined and the nature of the graphs of sine, cosine, and tangent (I'll assume you are familiar with their graphs in what follows).

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 $y \to - \setminus \infty$.