Problem 49 - 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.) $$ \lim _{h \rightarrow 0} \frac{1 /(5+h)-1 / 5}{h} $$

Step-by-Step Solution

Answer

Function: \( f(x) = 1/x \); Value \( c = 5 \).

Step 1: Identify the Given Limit as the Definition of Derivative

The given limit, \( \lim_{h \rightarrow 0} \frac{1/(5+h) - 1/5}{h} \), resembles the definition of the derivative. Specifically, it has the form \( f'(c) = \lim_{h \rightarrow 0} \frac{f(c+h) - f(c)}{h} \), where \( f(x) \) is a function, \( c \) is a specific value, and \( h \) is approaching 0.

Step 2: Determine the Function \( f(x) \)

In our limit, the term \( 1/(5+h) \) corresponds to \( f(c+h) \), which suggests that our function \( f(x) = 1/x \). This is because when \( h \) is 0, \( c + h \) becomes \( c \).

Step 3: Identify the Value \( c \)

From the limit \( \frac{1/(5+h) - 1/5}{h} \), it is clear that \( f(c) = 1/5 \). The value of \( x = c \) that makes \( f(x) \) equal to \( 1/5 \) is \( x = 5 \). Thus, we have \( c = 5 \).

Step 4: Confirm the Function and Value Values

We found \( f(x) = 1/x \) and \( c = 5 \). Substituting into the definition of the derivative at \( x = 5 \), \( f'(5) = \lim_{h \rightarrow 0} \frac{1/(5+h) - 1/5}{h} \), which matches exactly. Thus, the function and value are correct.

Key Concepts

Limit Definition of Derivative

Understanding the limit definition of a derivative is crucial for delving into calculus. A derivative, expressed as \( f'(c) \), measures how a function \( f(x) \) changes as its input \( x \) changes. The mathematical formula for a derivative at a specific point \( c \) can be written as:\[ f'(c) = \lim_{h \to 0} \frac{f(c+h) - f(c)}{h}\]Here, \( h \) represents a very small number, approaching zero. The expression \( f(c+h) - f(c) \) calculates the change in the function's value over a small change in \( x \). Dividing by \( h \) gives the rate of change, or the slope, of the function at \( c \). This is the foundation of understanding how derivatives apply to curves and rates of change in various fields.

Function Specification

To use the limit definition of derivative effectively, specifying the function \( f(x) \) is critical. In the given problem, the expression \( \frac{1}{5+h} \) suggests the function could be \( f(x) = \frac{1}{x} \). Electing this function, proceed to substitute \( x \) with \( c + h \), where \( c \) is a specific point on the function where we desire the derivative.

• In our example, substituting \( c + h \) in place of \( x \) leads to \( f(c+h) = \frac{1}{5+h} \).
• For deducing \( f(c) \), recognize it as the function value when \( h \) is zero, resulting in \( f(c) = \frac{1}{5} \).

Identifying the function allows you to apply calculus principles to analyze changes and behavior at precise points.

Mathematical Limits

Limits are the cornerstone of calculus, providing the means to understand behavior as variables approach certain values. A limit examines the value a function approaches as the input nears a specified point. In the exercise, the limit \( \lim_{h \rightarrow 0} \frac{1/(5+h)-1/5}{h} \) directly relates to the standard derivative form.

• This limit is telling us to evaluate how \( \frac{1}{5+h} - \frac{1}{5} \), divided by \( h \), behaves as \( h \) shrinks towards zero.
• The focus on \( h \rightarrow 0 \) is pivotal, confirming which function value we're approaching and calculating the rate of change at that instant.

Grasping how limits work helps decipher the small-scale behavior of functions, allowing for precise determination of derivatives and other calculus phenomena.