Does limit x tends to 1 sin inverse x exist? One thought says that since the limit x tends to 1+ does not exist therefore the limit does not exist. But other thought says that since the function is not defined at 1+, i.e. it is not in the domain of the function so we must not discuss limit over there and the limit is by default the limit x tends to 1-. Which thought is correct?
 A: $\text{Arcsin} \ x$ is defined on the interval $[-1,1]$, so you cannot take a right hand limit to $1$. So to answer your question directly, the latter thought you proposed is more accurate.
A: Are you talking about "$\lim_\limits{x\to1^{+}}\arcsin x$"? Since the function is undefined for $x>1$, this is a moot question. On the other hand, your both thoughts (some of which isn't quite clear) end up with the same conclusion that this limit doesn't exist. So I'm not sure what your question is, if there's no disagreement to resolve.
A: $\lim_{x\to 1}\sin^{-1}(x)$ does not exist, because $\sin^{-1}(x)$ is not defined for $x>1$. But $\lim_{x\to 1^-}\sin^{-1}(x)$ does exist, and is equal to $\sin^{-1}(1)=\pi/2$.
A: let's have an example :
$f(x) = \frac{x² - 25}{x-5}.$ Find $f(5)$.
As we know, if we find $f(5)$, the answer will be not defined.
But now let's find limit $x$ tends to $5$ for the same function.
When we will have an answer, we will come to know that the limiting value of the given function for $x=5$ approaches $10$.
So we can conclude that, AT A POINT, LIMITING VALUE OF FUNCTION MAY OR MAY NOT BE EXIST IRRESPECTIVE OF FUNCTION BEING DEFINED OR NOT DEFINED AT THAT POINT.
So limit value of $\sin^{-1} x$ exist and also limit exists, too if and only if we choose the value of $x$ from its domain.
