Problem 47 - Calculus Single Variable

Question

Specify a function \(f\) and a value \(c\) for which the given limit equals \(f^{\prime}(c) .\) (You need not evaluate the limit.) $$ \lim _{h \rightarrow 0} \frac{\sqrt{4+h}-2}{h} $$

Step-by-Step Solution

Answer

Function: \( f(x) = \sqrt{x} \), Evaluation Point: \( c = 4 \).

Step 1: Understanding the Expression

The given limit expression is \( \lim_{h \to 0} \frac{\sqrt{4+h} - 2}{h} \). This is in the form \( \frac{f(x+h) - f(x)}{h} \), which resembles the definition of the derivative of a function at a point.

Step 2: Identifying Function and Evaluation Point

Notice that inside the square root, \(4 + h\) suggests that the function involved is the square root function. Therefore, consider \( f(x) = \sqrt{x} \), and note that \(4 + h\) suggests evaluating this function at \(x = 4\). So, \( c = 4 \).

Step 3: Matching with Derivative Definition

The limit \( \lim_{h \to 0} \frac{f(c+h) - f(c)}{h} \) is the definition of the derivative \( f'(c) \). For our function \( f(x) = \sqrt{x} \) at \( c = 4 \), the limit expression exactly matches this derivative.

Key Concepts

Limit

The concept of a limit is fundamentally important in calculus. It helps us understand how a function behaves as its input approaches a particular point. In this context, a limit assesses the behavior of a function as it comes infinitesimally close to a specific value, but not necessarily when it reaches that value.
In the provided exercise, the limit expression is given as \( \lim_{h \to 0} \frac{\sqrt{4+h} - 2}{h} \). This represents how the expression \( \frac{\sqrt{4+h} - 2}{h} \) behaves as \( h \) approaches zero. Understanding limits is crucial because it forms the basis for finding the derivative of functions, which tells us how the functions change locally.
When evaluating this limit, we are essentially observing the behavior of the function \( \sqrt{x} \) near \( x = 4 \). This helps us dissect the function's behavior without directly reaching an undefined point. It's often the case in calculus that limits allow us to make sense of expressions that otherwise seem undefined by standard algebraic rules.

Square Root Function

The square root function, denoted as \( f(x) = \sqrt{x} \), is a fundamental function seen across mathematics. It takes a non-negative number and returns another number that, when multiplied by itself, results in the original number. In other words, \( \sqrt{x} \times \sqrt{x} = x \).

• For positive numbers, the square root is straightforward, and the domain encompasses all non-negative real numbers \((x \geq 0)\).
• The function \( \sqrt{x} \) is continuous and smooth for all its domain values, tending to increase slowly as \( x \) grows larger.

In our exercise, the function \( f(x) = \sqrt{x} \) evaluates at \( x = 4 \), allowing us to utilize the derivative definition. For this function, it is important to consider how \( \sqrt{4+h} \) changes as \( h \to 0 \), which combines with the concepts of limits and derivatives to provide insights into the rate of change for \( f(x) \) at specific points.
This subtle approach through the square root highlights the need for understanding incremental changes and approximations central to calculus.

Definition of Derivative

The definition of a derivative provides us with a precise way of calculating the rate at which a function changes at any given point. The derivative of a function \( f(x) \) at a point \( c \) is given by the limit
\[ f'(c) = \lim_{h \to 0} \frac{f(c+h) - f(c)}{h} \]
This expression symbolizes the average rate of change of the function over an interval of length \(h\), becoming precise as \(h\) becomes infinitesimally small.
When we evaluate the derivative at \( c \) in our problem, we're looking at \( \frac{\sqrt{4+h} - 2}{h} \), a classic formulation of this derivative process for the square root function, specifically when \( c = 4 \). This matches our task's limit expression, verifying that the function \( f(x) = \sqrt{x} \) and the evaluation \( c = 4 \) fit perfectly with the derivative definition.

• This calculation tells us exactly how fast \( \sqrt{x} \) is changing at \( x = 4 \), a central concern in calculus as it moves from algebraic equations to concepts concerning instantaneous rates and motions.

Understanding the derivative's definition helps in deeply appreciating how small changes in input cause changes in the function's output, a core concept in calculus.