Existence of limit involving arcsin Does $\mathrm{\lim\limits_{x\to \infty}arcsin\left(\frac{x+1}{x}\right)}$ exist?
Technically $x+1$ is always greater than $x$. Hence the limit should not exist.
However if we evaluate $\mathrm{\lim\limits_{x\to \infty}\frac{x+1}{x}}$ first and then plug it into $\arcsin$ we get $\arcsin(1)=\frac{\pi}2$.
Which one is it? "Does not exist" or $\frac{\pi}{2}$?

The function $\arcsin$ is not defined for any value greater than $1$, so the expression
$$\arcsin\left(\frac{x+1}x\right)$$
does not even make sense for $x>-1$. A limit when $x\to\infty$ is only defined if there is some interval $(a,\infty)$ contained in the domain of the function, and this is not the case at all.
