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$\lim_{x\to 1} [\sin^{-1} x]$ ; 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.

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

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  • $\begingroup$ why? by definition of limit, limit exists only when RHL = LHL, what is the basis that justifies this assumption? $\endgroup$ – Arjang Dec 20 '17 at 12:41
  • $\begingroup$ @Arjang The definition of limits from one side is absolutely consistent and used. en.wikipedia.org/wiki/Limit_of_a_function $\endgroup$ – user Dec 20 '17 at 13:07
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    $\begingroup$ Actually the RHS limit does exist in complex domain, but if we accept it does not exist, then LHS limit exists, RHS does not exist and therefore limit does not exist, the book is wrong, OP is correct to question the book. $\endgroup$ – Arjang Dec 20 '17 at 13:20
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    $\begingroup$ @Arjang Your definition of limit requires both sides. Other commonly found definitions don't and there's no absolute right or wrong here. It's a matter of conventions. $\endgroup$ – egreg Dec 20 '17 at 15:14
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    $\begingroup$ @Arjang You don’t, but I do. $\endgroup$ – egreg Dec 20 '17 at 23:23
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Yes, you are right to question the book, it is wrong, the limit does not exist, only RHS limit exists, but not the LHS limit ( In complex domain LHS limit also exists, but that will be part of Complex analysis course).

For your purposes the book is wrong, see https://en.wikipedia.org/wiki/One-sided_limit

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