Evaluating Limits: A Simple Example

In calculus, finding the limit of a function is a fundamental concept. It helps us understand the behavior of a function as it approaches a particular point. Let's explore how to evaluate the limit of the function ${\frac{x-9}{x^2-81}}$ as x approaches 9. This example will provide a clear, step-by-step approach that you can apply to similar problems. Understanding limits is crucial for grasping more advanced calculus topics, such as derivatives and integrals. So, let's dive in and break down this problem.

Understanding the Limit

Before we jump into the solution, let's quickly recap what a limit is. The limit of a function f(x) as x approaches a value c, written as ${\lim_{x \rightarrow c} f(x)}$, is the value that f(x) gets closer and closer to as x gets closer and closer to c. It's important to note that the limit doesn't necessarily equal the value of the function at x = c; in fact, the function might not even be defined at that point. In our case, we want to find out what happens to the function ${\frac{x-9}{x^2-81}}$ as x gets closer and closer to 9. This involves algebraic manipulation to simplify the expression and avoid any undefined forms, such as division by zero. Remember, limits are a foundational concept, so understanding them thoroughly is key to mastering calculus. This introductory understanding sets the stage for the practical steps we'll take to solve the problem. Evaluating limits is a core skill in calculus, forming the basis for understanding continuity, derivatives, and integrals. Keep this definition in mind as we work through the example.

The Problem: $\lim _{x \rightarrow 9} \frac{x-9}{x^2-81}$

Our task is to find the limit of the function ${\frac{x-9}{x^2-81}}$ as x approaches 9. At first glance, if we directly substitute x = 9 into the function, we get ${\frac{9-9}{9^2-81} = \frac{0}{0}}$, which is an indeterminate form. This means we can't directly evaluate the limit by substitution. Instead, we need to manipulate the expression algebraically to simplify it. The presence of the indeterminate form signals that there might be a common factor in the numerator and denominator that we can cancel out. Identifying and simplifying such expressions is a common technique in limit evaluations. The goal is to rewrite the function in a form where direct substitution is possible and yields a meaningful result. This initial assessment is crucial because it guides our approach to solving the problem. Recognizing the indeterminate form prompts us to look for algebraic simplifications that will allow us to evaluate the limit. Without this initial check, we might incorrectly assume that the limit does not exist or is undefined. Therefore, always start by attempting direct substitution and checking for indeterminate forms before proceeding with other methods.

Step-by-Step Solution

Here's how we can solve this limit problem:

1. Factor the denominator: The denominator ${x^2 - 81}$ is a difference of squares, which can be factored as ${(x - 9)(x + 9)}$. So, we can rewrite the function as:

   ${\frac{x-9}{x^2-81} = \frac{x-9}{(x-9)(x+9)}}$

2. Simplify the expression: Notice that we have a common factor of ${(x - 9)}$ in both the numerator and the denominator. We can cancel this factor out, as long as ${x \neq 9}$. This gives us:

   ${\frac{x-9}{(x-9)(x+9)} = \frac{1}{x+9}}$

3. Evaluate the limit: Now that we've simplified the expression, we can take the limit as x approaches 9:

   ${\lim_{x \rightarrow 9} \frac{1}{x+9} = \frac{1}{9+9} = \frac{1}{18}}$

Therefore, the limit of the function ${\frac{x-9}{x^2-81}}$ as x approaches 9 is ${\frac{1}{18}}$.

Conclusion

In summary, to evaluate the limit ${\lim _{x \rightarrow 9} \frac{x-9}{x^2-81}}$, we first recognized that direct substitution resulted in an indeterminate form. We then factored the denominator using the difference of squares, simplified the expression by canceling the common factor ${(x - 9)}$, and finally, evaluated the limit of the simplified expression. This process allowed us to find that the limit is ${\frac{1}{18}}$. This example demonstrates a common technique in calculus for evaluating limits: algebraic manipulation to simplify expressions and eliminate indeterminate forms. Understanding these techniques is essential for solving a wide range of limit problems. Remember to always check for indeterminate forms and look for ways to simplify the function before attempting to evaluate the limit directly. By mastering these skills, you'll be well-equipped to tackle more complex calculus problems. Keep practicing with different examples to solidify your understanding and build your problem-solving abilities. The ability to manipulate and simplify expressions is crucial for effectively evaluating limits. This step-by-step approach not only provides the solution but also reinforces the underlying principles of limit evaluation. Practice with similar problems will further enhance your skills and confidence in solving calculus problems.

To further your understanding of limits, you might find it helpful to explore additional resources. Check out Khan Academy's lessons on limits and continuity for more practice problems and detailed explanations.