Problem 33 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find \(\lim _{x \rightarrow 2}[f(x)-f(2)] /(x-2)\) for each given function \(f\). $$f(x)=\frac{1}{x}$$

Step-by-Step Solution

Answer

The limit is \( \frac{-1}{4} \).

Step 1: Understand the Limit Definition

The given problem involves finding the derivative of the function \( f(x) \) at \( x = 2 \) by using its limit definition. Specifically, the problem asks us to find: \[ \lim_{x \rightarrow 2} \frac{f(x)-f(2)}{x-2} \] This is the definition of the derivative \( f'(x) \) at \( x = 2 \).

Step 2: Evaluate f(2) for f(x)

Substitute \( x = 2 \) into the function \( f(x) = \frac{1}{x} \) to find \( f(2) \): \[ f(2) = \frac{1}{2} \]

Step 3: Substitute f(x) and f(2) into the Limit Expression

Substitute \( f(x) = \frac{1}{x} \) and \( f(2) = \frac{1}{2} \) into the limit expression: \[ \lim_{x \rightarrow 2} \frac{\frac{1}{x} - \frac{1}{2}}{x - 2} \]

Step 4: Simplify the Numerator

Combine the fractions in the numerator: \[ \frac{1}{x} - \frac{1}{2} = \frac{2 - x}{2x} \] To get this, find a common denominator (which is \( 2x \)) and rewrite the expression.

Step 5: Rewrite the Limit Expression

Now the limit expression becomes: \[ \lim_{x \rightarrow 2} \frac{\frac{2 - x}{2x}}{x - 2} \] Simplify further by rewriting: \[ \lim_{x \rightarrow 2} \frac{2 - x}{2x(x - 2)} \] which can be rearranged as: \[ \lim_{x \rightarrow 2} \frac{-(x - 2)}{2x(x - 2)} \]

Step 6: Cancel and Simplify Further

Cancel out the \( x - 2 \) terms in the numerator and the denominator: \[ \lim_{x \rightarrow 2} \frac{-1}{2x} \]

Step 7: Evaluate the Limit as x Approaches 2

Substitute \( x = 2 \) into the simplified expression: \[ \frac{-1}{2(2)} = \frac{-1}{4} \] Thus, the limit is \( \frac{-1}{4} \).

Key Concepts

Limit Definition

The limit definition of the derivative is a fundamental concept in calculus. It defines the derivative as the limit of the average rate of change of the function as the interval approaches zero. The limit expression we use is:

• \( \lim_{x \rightarrow c} \frac{f(x) - f(c)}{x - c} \)

This expression gives us the derivative of the function \( f(x) \) at a point \( c \). In our exercise, we are interested in determining \( f'(2) \) for the function \( f(x) = \frac{1}{x} \). This process involves evaluating the limit expression at \( x=2 \) to find the slope of the tangent line at that point.

Calculus

Calculus is the branch of mathematics dealing with derivatives and integrals, among other things. Derivatives are used to understand how a function changes at any point. In this context, we are using the limit definition of derivative to calculate \( f'(x) \). Calculus allows us to:

• Determine rates of change
• Understand motion, growth, or decline of quantities
• Find optimal solutions for various types of problems

Our exercise exemplifies using calculus to evaluate this limit and determine the instantaneous rate of change of the function \( f(x) = \frac{1}{x} \) at a specific point.

Function Evaluation

Function evaluation involves substituting a specific value of \( x \) into a function to determine what the function equals at that point. When solving the limit definition of the derivative, evaluating the function is necessary. In our step-by-step solution, we evaluated \( f(2) \) by substituting \( x = 2 \) into the given function:

• \( f(x) = \frac{1}{x} \)

This gives us \( f(2) = \frac{1}{2} \). By substituting \( f(x) \) and \( f(2) \) into the limit expression, we can simplify and solve for the derivative at a specific point.

Rational Functions

Rational functions are those written as the ratio of two polynomials. Our function \( f(x) = \frac{1}{x} \) is the simplest form of a rational function. When finding limits, these require techniques like factorization or common denominators to simplify. In our exercise, we needed to

• Simplify the expression in the numerator
• Use algebraic manipulation to cancel out terms where possible

This simplified the limit expression and allowed us to evaluate it by canceling common factors and substituting the limit point. Addressing rational functions often involves such simplifications to successfully compute limits and derivatives in calculus.