...on dividing both the numerator and the denominator by x and then putting $x=\infty$...
if you could just do that, then there wouldn't really be a need for ever using limits. $\infty$ is not a number (in standard analysis, that is), so you can't “put $x=\infty$”.
Instead, the whole idea of the limit is to put in ever larger finite values for $x$ and still get a result that's not only always finite, but actually converges towards some point (which we then call the limit). This does work for
because here, you always have $0<y<1$, so can always find a solution to $y = \sin t$, and because $y$ goes asymptotically to $1$, this† converges to a single point:
But it doesn't at all work for the limit you're asking about, because here $y>1$ for a finite $x$, and that means you don't actually ever get a solution at all. Thus there also can't be a limit.
†It only converges if you actually choose always the same solution, such as always the absolute-smallest one, which is what the $ⅹcsin$ function yields.