Boundedness of real functions: Survey of methods

Boundedness is a bit tricky since there is no direct procedure for determining it. However, there are certain approaches one can try.

1. Known boundedness. We know that some functions - notably some elementary functions - are bounded, for instance all constant functions, sine, cosine, and all inverse trig functions. If we combine such functions, we often obtain bounded functions, too. The following facts may be useful:

 • If f,g are bounded, then f + g, f − g, f⋅g, and f g are also bounded.
 • If f is bounded, then | f | is bounded and also the composition f (g) is bounded for any function g. In particular, any transformation of f is also bounded.
Note: If g is bounded, then f (g) need not be bounded. For instance, sin(x) is bounded, but cot(sin(x)) is not. (Indeed, values of sin(x) get arbitrarily close to 0, and cotangent tends to infinity for argument going to zero.)
 • If f is bounded and g does not approach 0 arbitrarily close (that is, inf(|g|) > 0, see separation from 0 in Continuity), then f /g is bounded.

2. Known unboundedness. We also know that some functions are unbounded, for instance polynomials, tangent and cotangent, exponential and logarithm etc. Again, some facts can be useful.

 • A sum/difference of a bounded and an unbounded function is again unbounded. Note that adding/subtracting two unbounded functions may yield a bounded function. For product and division anything can happen, for instance a product of two unbounded functions may be bounded. Example: x⋅(1/x) = 1.
 • Composition is unpleasant as well, a composition of two unbounded functions may yield a bounded function. For instance, 1/x is unbounded, so is (ex + 1), but when we substitute the latter into the former, we get 1/(ex + 1), which is a bounded function. Indeed, inf(ex + 1) = 1, so we are separated from zero, and 1 is a bounded function (constant), so the ratio is bounded by the last fact in part 1 above.

3. Theoretical tricks. These use knowledge of theory, we will just show some handy facts.

 • Functions continuous on bounded closed sets are bounded.
 • Functions bounded on a set are also bounded on all its subsets. This is usually used the other way around. When considering a function on a set, it might help to look at the some larger set on which the boundedness is easier to see. In particular, given an open set, it often pays off to first check on some larger closed set (typically by including endpoints), because then we can often use the previous handy fact.
 • If a function f is continuous on a set M which is an interval or a finite union of intervals, and all (one-sided) limits at endpoints of M are finite, then f is bounded on M.
 • If the range of g is a bounded closed set and f is a function continuous on that set, then the composition f (g) is bounded. In general, if the range of g is a set M that is an interval or a finite union of intervals, we can explore f on M as in the second fact in this part to determine boundedness of f (g).

4. Graphing. When all else fails, we can always sketch the graph of the given function using methods from Derivative - Theory - Graphing functions and then guess boundedness from the graph.

Example: Determine boundedness of

Solution: We should always start by determining the domain, here it is actually quite simple, the domain is all real numbers (check). We see that this function is a sum/difference of three terms, so we can explore each term separately and then try to put this information together.

The first term is a product. Sine is a bounded function, so no matter what we put into it, the outcome - here sin(ex) - is still bounded. The second term is a bit more tricky. Tangent is not bounded, but we are substituting a bounded function into it, which may remove the "bad" parts from consideration. Indeed, the range of cosine is the bounded closed interval [−1,1] on which tangent is continuous, therefore tan(cos(x)) is bounded. Finally, the product of two bounded functions is bounded.

The second term is a ratio of two functions. The function on the top is bounded, since it is just a transformation of a bounded function cosine (first horizontal shift and mirroring, then vertical shift and scaling). That's a good start, next we need to know whether the denominator can approach 0 arbitrarily close. However, x2 + 1 is always at least 1, so it cannot get any closer to 0. Consequently, the fraction is a bounded function.

The third term is a composition again. The outside function is not bounded, so we may have a problem. The inside function - arc cotangent - is bounded with range (0,π) and since logarithm is continuous everywhere, it is continuous also on this interval. Thus the decisive factor is the behaviour of logarithm at the ends of this range. There is no problem at π, but we know that as the argument of logarithm approaches zero, the logarithm blows up to negative infinity. Thus this third term is not bounded.

Sum of the first two bounded terms is bounded, so we have a bounded expression from which we subtract an unbounded one. Consequently, the given function is unbounded.

 

For further examples we refer to Solved Problems.

Symmetry
Back to Methods Survey - Basic properties of real functions