2

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

Siddharth Garg
  • 189

2 Answers2

4

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

user
  • 154,566
  • why? by definition of limit, limit exists only when RHL = LHL, what is the basis that justifies this assumption? – jimjim Dec 20 '17 at 12:41
  • @Arjang The definition of limits from one side is absolutely consistent and used. https://en.wikipedia.org/wiki/Limit_of_a_function – user Dec 20 '17 at 13:07
  • that is one sided limit, it is not same thing as limit https://en.wikipedia.org/wiki/One-sided_limit – jimjim Dec 20 '17 at 13:15
  • @Arjang that is why the limit needs to be assumed as x\to 1^-, for arcsinx the RHS limit is meaningless – user Dec 20 '17 at 13:16
  • 1
    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. – jimjim Dec 20 '17 at 13:20
  • Often also for $\log x$ usually $x\to0$ is indicated and that obviously means $x\to0^+$ since the limit for $x\to 0^-$ is meaningless. – user Dec 20 '17 at 13:25
  • so they abuse the notation, OP correctly stated that limit exists when LHS limit = RHS limit. People being lazy and not exact does not indicate that limit exists. Although it does exist in complex domain. – jimjim Dec 20 '17 at 13:30
  • What I claim is that in the case of $arcsin x$ you can't consider the RHS limit in 1 because the domain is [-1,1] thus the RHS limit can't be defined at all. Thus when you write $x\to 1$ it is to be intended$ x\to 1^-$. – user Dec 20 '17 at 13:33
  • Cant do that, because $x \to 1$ means both $x \to 1^+ = x \to 1^-$, it is not consistent for it to also sometimes only mean $x \to 1^+$ and some other time to only mean $x \to 1^-$ exclusively. – jimjim Dec 20 '17 at 13:42
  • @Arjang I've understood your point but in this case I think that the notation is cleraly indicating that the limit has to be considered at LHS. For instance, It's a different case from the limit $\frac{1}{x}$ for $x\to0$. – user Dec 20 '17 at 14:11
  • 1
    @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. – egreg Dec 20 '17 at 15:14
  • @egreg : That is not my definition of limit, it is how limit is defined. The OP mentioned that himself as well. I dont know any definition of limit were the limit is defined and $ LHL$neq RHL$. – jimjim Dec 20 '17 at 20:20
  • 1
    @Arjang You don’t, but I do. – egreg Dec 20 '17 at 23:23
  • @egreg I don't understand how that definition can be consistent with the common definition of limit. Of course you are welcome to your own definition and extensions as long as they are consistent, – jimjim Dec 21 '17 at 04:57
0

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

jimjim
  • 9,675