Problem 18 - Precalculus: Mathematical for Calculus

Question

Evaluate the limit, if it exists. $$\lim _{h \rightarrow 0} \frac{(3+h)^{-1}-3^{-1}}{h}$$

Step-by-Step Solution

Answer

The limit is \( \frac{-1}{9} \).

Step 1: Substitute and Simplify the Expression

Our goal is to find \( \lim _{h \rightarrow 0} \frac{(3+h)^{-1}-3^{-1}}{h} \). Let's substitute the inverse values: \( (3+h)^{-1} = \frac{1}{3+h} \) and \( 3^{-1} = \frac{1}{3} \). The expression becomes \( \lim _{h \rightarrow 0} \frac{\frac{1}{3+h} - \frac{1}{3}}{h} \).

Step 2: Find a Common Denominator

To combine \( \frac{1}{3+h} \) and \( \frac{1}{3} \), we need a common denominator, which is \( 3(3+h) \). Rewrite the expression: \[ \frac{1}{3+h} - \frac{1}{3} = \frac{3 - (3+h)}{3(3+h)} = \frac{-h}{3(3+h)}. \]

Step 3: Simplify the Fraction

Substitute the expression from Step 2 back into the limit: \( \lim _{h \rightarrow 0} \frac{\frac{-h}{3(3+h)}}{h} \). Simplify the fraction: \[ = \lim _{h \rightarrow 0} \frac{-h}{h \cdot 3(3+h)} = \lim _{h \rightarrow 0} \frac{-1}{3(3+h)}. \]

Step 4: Evaluate the Limit

Now, find the limit by substituting \( h = 0 \) in the simplified expression: \[ \lim _{h \rightarrow 0} \frac{-1}{3(3+h)} = \frac{-1}{3 \cdot 3} = \frac{-1}{9}. \] Hence, the limit exists and equals \( \frac{-1}{9} \).

Key Concepts

Understanding Calculus

Calculus is a branch of mathematics that deals with rates of change and the accumulation of quantities. It is primarily concerned with two fundamental concepts: differentiation and integration. Differentiation focuses on finding the rate of change of a quantity, while integration is concerned with finding the total accumulation of a quantity over an interval.
In calculus, we often deal with the concept of limits. Limits help us understand the behavior of functions as they approach a specific input. This is crucial for understanding how functions change dynamically and form the basis for more complex operations like derivatives and integrals.
In this exercise, we use limits to evaluate how a function behaves as the parameter \( h \) approaches zero. We are looking at a specific form of a function where calculus helps determine a precise value for this approach. Calculus provides a powerful set of tools for analyzing and understanding changes in mathematical functions, which is essential for science, engineering, economics, and many other fields.

Limit Evaluation Techniques

Limit evaluation is a method to determine the value that a function approaches as the input approaches a certain point. Evaluating limits involves manipulating an expression in such a way that allows us to safely take the limit without indeterminate forms like \( \frac{0}{0} \).
In the given problem, you must simplify the expression to make the limit evident. This typically involves algebraic manipulation, such as factoring, canceling out terms, or combining fractions. An important step in this exercise is converting the expression into a form with a single limitable component.

• The substitution step replaced complex fractions with more manageable expressions.
• Finding a common denominator allowed simplifying the expressions to a more recognizable form for limit evaluation.
• After simplification, substituting \( h = 0 \) into the expression determined the limit.

Effectively evaluating limits requires understanding both the algebraic and conceptual tools available in calculus. This practice solidifies comprehension of continuity and change within functions.

The Role of Common Denominators

A common denominator is a shared multiple of the denominators of two or more fractions. It allows us to combine and compare fractions effectively. This is especially useful in limit problems where multiple fractional expressions are involved.
In the problem, the denominators \( 3+h \) and \( 3 \) need to be combined. To do this, we find the common denominator \( 3(3+h) \). By rewriting each fraction with this denominator, we subtract one fraction from the other seamlessly.

• This process shows how to join complex fractions with different denominators into a single fraction.
• It helps to express the difference between the fractions, paving the way for further simplifications.
• Ultimately, using a common denominator is key to simplifying many calculus problems involving fractions, making them easier to analyze and solve.

Understanding this technique is fundamental in calculus, as it often simplifies the process of limit evaluation and other algebraic manipulations.