Question

Determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false. If \(f\) is continuous on \([0, \infty)\) and \(\lim _{x \rightarrow \infty} f(x)=0\), then \(\int_{0}^{\infty} f(x) d x\) converges

Short answer

The statement is false. An example is the function \(f(x)=1/x\). This function is continuous on (0, \infty) and as \(x\) becomes very large, \(1/x\) gets very close to 0. However, the integral from 1 to infinity is not finite. Hence, the function \(f(x)=1/x\) is a counterexample to the statement.

Step-by-step solution

Verifying the statement

The given statement is not always true. Though it might seem intuitive that if a function approaches 0 as x approaches infinity the integral from 0 to infinity would converge, but it is not always the case. For a function to have a finite integral, it's not just important that the function approaches to 0, but it 'matters how quickly' it approaches to 0. There could be functions which approach 0 very slowly resulting in a divergent integral.

Counterexample

Consider for example the function \(f(x)=1/x\). This function is continuous on the interval \((0, \infty)\) and as x becomes very large, \(f(x)\) gets very close to 0. However, the area under the curve is not finite. \nThe integral \(\int_{1}^{\infty} 1/x dx\) evaluates to \(ln|x|\) from 1 to infinity, which goes to infinity. Hence, though the function \(f(x)=1/x\) is continuous on \([0, \infty)\) and approaches 0 as x approaches infinity, the integral does not converge.

Conclusion

So, the statement provided is false. A counterexample was found and the integral of the function from 0 to infinity doesn't always converge simply because it approaches 0 as x tends to infinity.

Key concepts

Continuous Functions

A continuous function is one that does not have any holes, breaks, or jumps within its domain. In mathematical terms, a function f is said to be continuous at a point x=a if the following three conditions are met:

• The function f(a) is defined.
• The limit of f(x) as x approaches a exists.
• The limit of f(x) as x approaches a is equal to f(a).

If these conditions hold true for every point within an interval, then the function is continuous over that interval. However, continuity alone does not guarantee that the integral of a function over an infinite interval will converge. The rate at which the function approaches a limiting value, or how it behaves overall, can have a significant impact, which leads us to concepts like improper integrals and limits at infinity.

Improper Integrals

An improper integral is an integral that either has infinite limits of integration or an integrand that approaches infinity at one or more points within the range of integration. To evaluate an improper integral, we typically convert it into a limit problem where we can apply the rules of calculus.

For instance, consider the integral of f(x) from 0 to infinity. What we’re actually doing here is taking the limit as b approaches infinity of the integral from 0 to b of f(x). If this limit exists and is finite, we say the integral converges; otherwise, it diverges. The key factor in determining convergence is not just where the function goes as x approaches infinity, but how it gets there, meaning the manner in which it approaches the limit.

Limits at Infinity

The concept of limits at infinity refers to what value a function approaches as its input grows without bound. Mathematically,

\( \lim_{x \rightarrow \infty} f(x) \)

represents the value that f(x) gets arbitrarily close to as x becomes very large. If f(x) approaches a specific number, this limit is said to exist. Nonetheless, a function may have a limit at infinity, such as zero, and still have an integral that diverges. This somewhat surprising outcome occurs because it is not only the limit that matters, but also the rate at which the function approaches this limit. For example, the function f(x) = 1/x approaches zero as x approaches infinity, but the area under f(x) over the interval from 1 to infinity increases without bound, and thus the integral of f(x) from 1 to infinity diverges. This illustrates an important distinction: a function can approach a limit at infinity slowly enough that the total accumulation (area under the curve) is infinite, leading to a divergent improper integral.