Problem 28 - Calculus Early Transcendentals

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: Substitution and Simplification

Start by writing the expression you have to evaluate: \( \lim _{h \rightarrow 0} \frac{(3+h)^{-1}-3^{-1}}{h} \). Since direct substitution of \( h = 0 \) results in a \( \frac{0}{0} \) form, we need to simplify it. So, substitute \( a = 3+h \) to then write \( (3+h)^{-1} = a^{-1} \). Simplify the expression to get to: \[ \lim_{h \to 0} \frac{1}{h} \left( \frac{1}{3+h} - \frac{1}{3} \right) \].

Step 2: Finding a Common Denominator

The next step is to combine the two fractions into one. To do so, find a common denominator for the fractions: \( \frac{1}{3+h} \) and \( \frac{1}{3} \). The common denominator is \((3+h)(3)\). Rewrite the expression:\[ \frac{1}{3+h} - \frac{1}{3} = \frac{3 - (3+h)}{3(3+h)} = \frac{3 - 3 - h}{3(3+h)} = \frac{-h}{3(3+h)} \].

Step 3: Simplify the Expression

Replace the single fraction back into the limit expression:\[ \frac{1}{h} \cdot \frac{-h}{3(3+h)} = \frac{-1}{3(3+h)} \].Cancel the \( h \) in the numerator and denominator.

Step 4: Evaluate the Limit

Now that the expression is simplified, substitute \( h = 0 \) into the expression.\[ \lim_{h \to 0} \frac{-1}{3(3+h)} = \frac{-1}{3 \cdot 3} = \frac{-1}{9} \].Thus, the limit evaluates to \( \frac{-1}{9} \).

Key Concepts

Limits in Calculus

Calculus is a mathematical field that studies how quantities change, and limits are a foundational concept in this field. A limit helps us understand the behavior of a function as its input gets closer to a certain value. For example, the notation \( \lim_{h \to 0} \) asks what value a function approaches as \( h \) gets arbitrarily close to zero. When calculating limits, we often encounter expressions that become difficult to evaluate directly. In these cases, simplifying the expression or using algebraic tricks can help. The goal is to transform the expression into a form where substitution becomes possible. If we successfully simplify it, we can find the limit value easily.Limits are used not only in calculus to define derivatives but also in other areas such as sequences, series, and even integrals. Understanding limits is crucial for mastering further calculus topics.

Derivatives

The concept of derivatives is closely related to limits. In calculus, a derivative represents the rate of change of a function with respect to one of its variables. It gives us information on how a function behaves as its input changes.The formal definition of a derivative uses a limit. If \( f(x) \) is a function, the derivative \( f'(x) \) is calculated by:\[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}\]This expression looks a lot like the limit we evaluated in the given exercise. It shows how derivative calculations can often involve working through similar algebraic simplifications.Derivatives have many practical applications, from determining velocity in physics to analyzing marginal costs in economics. It is vital to understand how derivatives function, not just in theory but in real-world applications.

Indeterminate Forms

During limit evaluation, we may encounter forms like \( \frac{0}{0} \), which are known as indeterminate forms. These forms don't directly tell us the limit's value, and we need to apply techniques like factoring, rationalizing, or using algebraic identities to resolve them.In the provided solution, when substituting \( h = 0 \) in the limit expression, \( \frac{(3+h)^{-1}-3^{-1}}{h} \) initially resulted in the indeterminate form \( \frac{0}{0} \). To find a meaningful limit, we simplified the algebraic expression step-by-step, getting rid of the indeterminate form.Understanding how to manage and manipulate these indeterminate expressions is critical. Mastery of this concept not only aids in computing limits but is also instrumental in understanding continuity and differentiable functions.

Rational Functions

Rational functions are fractions where both the numerator and the denominator are polynomials. In calculus, they often come up when solving limit problems or other calculus-related tasks.Analyzing these functions involves understanding their behavior, especially as the variable approaches certain critical points, such as where the denominator becomes zero. Simplifying rational expressions often involves factorization, finding common denominators, or performing long division.In the example exercise, we dealt with the rational function \( \frac{1}{h} \left( \frac{1}{3+h} - \frac{1}{3} \right) \). We found a common denominator and simplified the expression, allowing us to evaluate the limit correctly.Recognizing patterns in rational functions and applying the correct algebraic techniques simplifies their limits or derivatives, making these functions much more manageable in calculus.