This question might seem a bit rudimentary, but I've been fumbling around with this same problem all week so I'm losing the forest for the trees a little bit. I won't get into the whole problem, so I'll give a for-instance instead.

Consider a function f:R->R where f(x) = sqrt(x) over the interval [1,inf). Say we were unsure if f has finite values over this interval. Obviously for a simple function like this, all finite values of x>=1 have a finite value for f(x), ie, we can take the square root of all finite positive real numbers. But how could you prove this to be true?

My first thought is to look at its end behavior, but f->inf as x->inf, so that isn't terribly useful. If we were taking an integral that could tell us its divergent, but it doesn't actually tell us anything about the function itself. Would you have to prove a lack of vertical asymptotes over the interval? Surely there's a better way of doing this. The problem itself is quite large and this is only one very small piece so my brain is kinda fried, you'll have to excuse me if this is a dumb question, I'm still an underclassman