# Solve the following limit.

$\lim_{x\to 1} [\sin^{-1} x]$ ; where [.] is the 'Greatest Integer Function'.

The left hand limit will be $[π/2]$ = $1$. But how can there be a right hand limit (as $'x'$ can't take values greater than $1$)? The answer in my textbook is given as $1$. But how can the limit exist when there is no right hand limit because for a limit to exist, LHL should be equal to the RHL.

• RHL does exist, just ask wolfram alpha for values of x>1 – Arjang Dec 20 '17 at 12:36
• But even the domain of arcsin is [-1,1]. – Siddharth Garg Dec 20 '17 at 12:37
• so why there is a value for arcsin of 2? – Arjang Dec 20 '17 at 12:39
• – Siddharth Garg Dec 20 '17 at 12:50
• m.wolframalpha.com/input/?i=arcsin+2 – Arjang Dec 20 '17 at 13:01

Note that $\arcsin x$ is only defined for $x\in [-1,1]$ thus the limit is to be assumed as $x\to1^-$.