Question

Suppose \(f\) is defined and differentiable for every \(x>0\), and \(f^{\prime}(x) \rightarrow 0\) as \(x \rightarrow+\infty\). Put \(g(x)=f(x+1)-f(x)\). Prove that \(g(x) \rightarrow 0\) as \(x \rightarrow+\infty\).

Short answer

Question: Show that if a function \(f(x)\) is differentiable for every \(x > 0\) and its derivative approaches \(0\) as \(x\) goes to positive infinity, then for the function \(g(x) = f(x+1) - f(x)\), \(g(x)\) also approaches \(0\) as \(x\) goes to positive infinity. Short Answer: By applying the Mean Value Theorem to the function \(f(x)\) in the interval \([x, x+1]\), we can show that there exists a number \(c\) in the interval \((x, x+1)\) such that \(f'(c) = g(x)\). Since we are given that the limit of \(f'(x)\) is \(0\) as \(x\) goes to positive infinity, we can conclude that this also holds for the limit of \(g(x)\), i.e., \(\lim_{x \rightarrow +\infty} g(x) = 0\). Therefore, \(g(x)\) approaches \(0\) as \(x\) goes to positive infinity.

Step-by-step solution

State the Mean Value Theorem (MVT)

The Mean Value Theorem states that if a function \(h(x)\) is continuous on the closed interval \([a, b]\) and differentiable on the open interval \((a, b)\), then there exists a number \(c\) in the interval \((a, b)\) such that $$h'(c) = \frac{h(b) - h(a)}{b - a}.$$ We will use the MVT for the function \(f(x)\) in the interval \([x, x+1]\).

Apply the Mean Value Theorem to \(f(x)\) and \(g(x)\)

Since \(f(x)\) is differentiable for every \(x>0\), it is also continuous for every \(x > 0\) (differentiability implies continuity). Therefore, by the MVT, there exists a number \(c\) in the interval \((x, x+1)\) such that $$f'(c) = \frac{f(x+1) - f(x)}{x+1 - x}.$$ Note that the right-hand side of the equation is equal to \(g(x)\). Thus, we have $$f'(c) = g(x).$$

Analyze the limit of \(f'(c)\)

We are given that \(f'(x)\) approaches \(0\) as \(x\) goes to positive infinity, i.e., $$\lim_{x \rightarrow +\infty} f'(x) = 0.$$ Since \(c\) is in the interval \((x, x+1)\), as \(x\) goes to positive infinity, \(c\) also goes to positive infinity. Therefore, $$\lim_{c \rightarrow +\infty} f'(c) = 0.$$

Find the limit of \(g(x)\)

Now we know that $$\lim_{c \rightarrow +\infty} f'(c) = \lim_{c \rightarrow +\infty} g(x).$$ Since the limit of \(f'(c)\) as \(c\) goes to positive infinity equals \(0\), we can conclude that $$\lim_{x \rightarrow +\infty} g(x) = 0.$$ This means that \(g(x)\) approaches \(0\) as \(x\) goes to positive infinity, as required to show.

Key concepts

Differentiable Functions

Understanding differentiable functions is crucial when exploring the behavior of equations in calculus. A function is said to be differentiable over an interval if the derivative of that function exists at each point in the interval. More formally, for any function f(x), if the limit as h approaches zero of (f(x+h) - f(x))/h exists, then f is differentiable at the point x. This limit, if it exists, represents the function's derivative at x, denoted as f'(x).

According to the exercise, the function f is defined and differentiable for every x>0, which means that f'(x) exists for all those values. Differentiability implies that a function is locally linear, and its graph has a tangent at each point in its domain. Importantly, if a function is differentiable at a point, it is also continuous at that point, although the converse is not always true. In summary, differentiability assures us that f is smooth and has no sharp corners or breaks for x>0.

Limits of Functions

In the realm of calculus, the concept of limits helps to describe the behavior of functions as the input approaches a certain value. Limits are essential in defining both derivatives and integrals—the cornerstone concepts of calculus. The limit of a function f(x) as x approaches a value a is the value that f(x) gets closer to as x gets closer to a.

For the given exercise, we're particularly interested in the behavior of f'(x) as x approaches infinity. Here, it is stated that f'(x) approaches 0, written mathematically as \(\text{lim}_{x \rightarrow +\infty} f'(x) = 0\). This expression denotes that as the input value grows without bound, the rate of change of the function, which is depicted by its derivative, gets closer and closer to zero. This information is a key component that leads us to understand the behavior of the function g(x) as x also grows without bound.

Continuous Functions

Continuity is a property of functions that is intuitive at its core: A function f(x) is continuous at a point x=a if there is no gap, jump, or disruption in the graph of the function at that point. For f to be continuous at a, the function must be defined at a, the limit as x approaches a must exist, and the limit as x approaches a must equal f(a). A function that is continuous across an interval has a graph that can be traced without lifting the pencil.

Related to the textbook exercise, we assume that f(x) is not only differentiable but also continuous for every x>0. This is significant because the Mean Value Theorem, which is applied in the solution, requires the function f to be continuous on the closed interval [x, x+1]. The continuity of f is essential to ensure that there are no gaps or jumps that would otherwise invalidate the conclusion obtained by applying the theorem in deducing the behavior of g(x).