Problem: Investigate boundedness of the following function:
Solution:
This function is a sum of three terms, so we start by investigating each in
turn. But first we determine the domain of this function. The first term
requires that
D( f ) = (0,1).
The first term is a function with domain
The second term is a ratio. The numerator is a product of arcsin,
which is a bounded function, and the exponential, which is not a bounded
function as a whole, but it is continuous and hence also bounded on the
bounded closed interval
Now we need to check on the denominator. If it is separated from zero, we
have a bounded fraction. Unfortunately, on the interval
but we have some at the other end:
Thus this second term is not bounded on
The third term needs to be investigated, even though we lost boundedness in the second term, since if this one also came out unbounded, it could onceivably happen that those two "unboundednesses" would cancel each other out.
The third term is a ratio again. Is the numerator bounded? The function
x is continuous on
Indeed, if the numerator is bounded, then so is the whole fraction, since
we already observed that the denominator is separated from zero, precisely,
So here we go. Just to be on the safe side we look at the right end, too, although we already saw that no problem arises there.
What is really interesting is the left end, but this is a textbook example:
Since we've got finite limits at the endpoints, the numerator is bounded and therefore (due to the denominator separated from zero) also the whole fraction is bounded.
Now we come to the conclusion. The first and the third term is bounded, so these two cannot cancel the unboundedness we found in the second term. Therefore the given function is not bounded.