Problem 44 - Calculus Early Transcendentals: Pearson New International Edition

Question

Use the Mean Value Theorem to prove that $$ \lim _{x \rightarrow \infty}(\sqrt{x+2}-\sqrt{x})=0 $$

Step-by-Step Solution

Answer

As \( x \to \infty \), \( \sqrt{x+2} - \sqrt{x} \to 0 \) by MVT.

Step 1: State the Mean Value Theorem

The Mean Value Theorem (MVT) states that for a differentiable function \( f \) on the open interval \( (a, b) \), there exists a number \( c \) in \( (a, b) \) such that \( f'(c) = \frac{f(b) - f(a)}{b-a} \).

Step 2: Define the Function

Let \( f(x) = \sqrt{x} \). This function is differentiable on \( (x, x+2) \) for \( x > 0 \).

Step 3: Apply MVT to the Interval

According to the MVT, for the interval \( [x, x+2] \), there exists some \( c \in (x, x+2) \) such that \[f'(c) = \frac{f(x+2) - f(x)}{2} = \frac{\sqrt{x+2} - \sqrt{x}}{2}.\] Thus, \( f'(c) = \frac{1}{2}\sqrt{c} \), using the derivative \( f'(x) = \frac{1}{2\sqrt{x}} \).

Step 4: Equation from MVT

Rearrange to find \( \sqrt{x+2} - \sqrt{x} \). Using MVT, we have:\[\sqrt{x+2} - \sqrt{x} = \frac{2}{2\sqrt{c}} = \frac{1}{\sqrt{c}}.\]

Step 5: Evaluate the Limit

As \( x \to \infty \), \( c \to \infty \) because \( c \in (x, x+2) \). Hence, \( \frac{1}{\sqrt{c}} \to 0 \) as \( c \to \infty \). Thus, \[\lim_{x \to \infty} (\sqrt{x+2} - \sqrt{x}) = 0.\]

Key Concepts

Differentiable Functions

To understand how the Mean Value Theorem helps in finding limits, let's delve into the idea of a differentiable function. A differentiable function is one that has a derivative at every point in its domain. This means you can find its rate of change or slope at any specific point. For example, if you look at the function \( f(x) = \sqrt{x} \), it’s a differentiable function because you can compute its derivative, \( f'(x) = \frac{1}{2\sqrt{x}} \).

• The gradient or slope gives insight into how the function behaves at or around that point.
• Continuity is key: a differentiable function must be continuous, so there are no sudden jumps or holes.

In the context of our problem, understanding that \( \sqrt{x} \) is differentiable on \( (x, x+2) \) is crucial. It permits the use of the Mean Value Theorem, which relies on the behavior of the derivative to relate differences in function values to the derivative at some, as yet unknown, point \( c \).

Limit of a Function

A limit describes where a function is heading as the input approaches a certain value or infinity. In simpler terms, it helps us understand the long-run behavior of functions. In our specific problem, we are interested in the limit as \( x \to \infty \) for the expression \( \sqrt{x+2} - \sqrt{x} \). This seemingly small difference between two square roots helps illustrate the power of the Mean Value Theorem when computing limits.

• The goal of limits is to capture the idea of 'approaching' or 'tending towards'.
• They can be used to find long-term trends or behaviors of functions as they go towards infinity.

As \( x \to \infty \), the \( \frac{1}{\sqrt{c}} \) from our MVT conclusion diminishes to zero, informing us of the behavior of our original expression. This manipulation is a great way to illustrate practical applications of calculus, turning a complex expression into an approachable one using limits.

Application of Calculus

Calculus is a powerful tool in mathematics to solve intricate problems, often involving change and accumulation. Its application is broad, ranging from verifying the behavior of a function at points, maximizing profit, or understanding motion, just to name a few. Here, we see calculus employed in using the Mean Value Theorem to prove a limit.

The Mean Value Theorem (MVT) essentially tells us there is at least one point where the average rate of change (slope) over an interval equals the actual derivative (instantaneous rate of change) at some point in the interval.

• MVT provides insights that other tools might not, by bridging the gap between average and instantaneous change.
• By applying MVT, the limit \( \lim_{x \to \infty} (\sqrt{x+2} - \sqrt{x}) = 0 \) is efficiently proven, showcasing calculus's ability to simplify complex questions.

Real-world applications of calculus, like this exercise, demonstrate its effectiveness in breaking down problems to highlight their logical conclusions.