Problem 31 - Precalculus: Mathematics for Calculus

Question

Finding Limits 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 Rewrite the Expression

The given limit expression is \( \lim_{h \rightarrow 0} \frac{(3+h)^{-1}-3^{-1}}{h} \). First, express the fractions in terms of a common denominator. Rewrite \( (3+h)^{-1} \) as \( \frac{1}{3+h} \) and \( 3^{-1} \) as \( \frac{1}{3} \). The expression becomes \( \lim_{h \rightarrow 0} \frac{\frac{1}{3+h} - \frac{1}{3}}{h} \).

Step 2: Combine the Numerator

Combine the fractions in the numerator: \( \frac{1}{3+h} - \frac{1}{3} = \frac{3 - (3+h)}{3(3+h)} = \frac{-h}{3(3+h)} \). Therefore, our limit becomes \( \lim_{h \rightarrow 0} \frac{-h}{h \cdot 3(3+h)} \).

Step 3: Simplify the Expression

Cancel the \( h \) in the numerator and denominator: \( \frac{-h}{h \cdot 3(3+h)} = \frac{-1}{3(3+h)} \). So the limit now is \( \lim_{h \rightarrow 0} \frac{-1}{3(3+h)} \).

Step 4: Evaluate the Limit

Substitute \( h = 0 \) into the simplified expression. This gives \( \frac{-1}{3(3+0)} = \frac{-1}{9} \). Therefore, \( \lim_{h \rightarrow 0} \frac{-1}{3(3+h)} = \frac{-1}{9} \).

Key Concepts

Limit Evaluation

When approaching the concept of limit evaluation, think of it as the tool that helps us understand the behavior of a function as the inputs approach a certain value. With limits, we essentially ask: "What value does the function get closer to, even if it never actually reaches that point?" This is especially useful for understanding scenarios where substituting directly into a function results in an indeterminate form, like division by zero.

In our exercise, the goal is to find the limit as the variable \( h \) approaches zero in the expression \( \lim _{h \rightarrow 0} \frac{(3+h)^{-1}-3^{-1}}{h} \). This kind of limit is common when dealing with derivatives, as it represents the slope or rate of change of a function at a specific point.

The solution involves understanding what happens near the point of interest (\( h = 0 \)) without needing the function to be defined exactly at that point. This requires combining algebraic manipulation with logical reasoning to find the precise value the expression approaches as \( h \) gets infinitesimally small.

Fraction Simplification

Simplifying fractions is an essential skill in mathematics, as it often makes complex expressions easier to work with. In the context of finding limits, simplifying fractions helps us deal with potentially undefined expressions by removing common terms.

In our exercise, we are given \( \lim_{h \rightarrow 0} \frac{\frac{1}{3+h} - \frac{1}{3}}{h} \). This involves combining the two fractions in the numerator. First, we express each term with a common denominator. This means rewriting \( \frac{1}{3+h} \) and \( \frac{1}{3} \) as a single fraction:\[\frac{1}{3+h} - \frac{1}{3} = \frac{3 - (3+h)}{3(3+h)} = \frac{-h}{3(3+h)}.\]
Now, the complex fraction is simpler, allowing us to find common terms. We cancel the \( h \) terms from the numerator and denominator, which makes the expression \( \frac{-1}{3(3+h)} \).

This step is crucial because it transforms the original expression into a form that can be easily evaluated when we apply the limit, which eliminates any indeterminate forms like division by zero.

Substitution in Limits

Substitution in limits refers to the technique of directly replacing the variable approaching a limit with the value it approaches, once the expression is simplified and no longer indeterminate.

In the final step of our exercise, we need to find \( \lim_{h \rightarrow 0} \frac{-1}{3(3+h)} \). At this point, the expression is no longer complicated or undefined when \( h \) is zero. So, we can substitute \( h = 0 \) into the expression, resulting in:\[ \frac{-1}{3(3+0)} = \frac{-1}{9}. \]
This confirms that the expression's limit as \( h \) approaches zero is \( \frac{-1}{9} \).

Substituting in limits is a powerful tool because, after appropriate simplification, it enables us to attain the exact value that the function approaches without any more complex evaluation needed. Always ensure the expression is valid for substitution at the specific point the limit approaches.