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 $\frac{\pi}{2}$?
It is often true that
$$\lim_{x\to\infty}f(g(x))\ne f\left(\lim_{x\to\infty}g(x) \right)$$
You cannot just pass a limit through a function. Many such examples are given in a typical first-year calculus class. You have hit on one of those cases. Your limit does not exist, for the reason you gave.
The answer depends on which definition of the $arcsin$ function the OP had in mind. If it is defined on the real numbers, then -as all the other responders have shown- the function is not defined for arguments larger than $1$. Consequently the limit does not exist.
On the other hand, it is well known that the definition of the $arcsin$ function can be extended to the complex plane. In that case there is no problem in evaluating the function value for argument $1 + 1/x$, with $x$ large. The result is (in lowest order approximation): $\pi/2 - i\sqrt{2/x}$. Taking the limit of $x$ to infinity yields the correct result: $\pi/2$.
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.
It doesn't exist, since a limit approaches an (unexisting) value. If the function does not exist for approaching values the limit can not be calculated.
