Solving Limits In Calculus Finding The Limit Of (x^2+x-12)/(x^2-3x) As X Approaches 3

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Introduction

In the realm of calculus, understanding limits is a foundational concept. Limits allow us to analyze the behavior of functions as they approach specific points, even if the function is not defined at those points. This article delves into the process of evaluating a specific limit: lim⁑xβ†’3x2+xβˆ’12x2βˆ’3x\lim_{x \rightarrow 3} \frac{x^2+x-12}{x^2-3x}. We will explore the techniques involved in finding this limit, providing a comprehensive understanding for readers. This exploration not only reinforces the fundamental principles of limits but also showcases the practical application of algebraic manipulation in resolving indeterminate forms.

The Concept of Limits in Calculus

In calculus, the concept of a limit describes the value that a function approaches as the input (or independent variable) approaches some value. This β€œsome value” might be a specific number, infinity, or negative infinity. In simpler terms, a limit tells us where the function is β€œheading.” Formally, the limit of f(x)f(x) as xx approaches cc is denoted as:

lim⁑xβ†’cf(x)=L\lim_{x \rightarrow c} f(x) = L

This notation indicates that as xx gets arbitrarily close to cc, the function f(x)f(x) gets arbitrarily close to LL. The critical aspect here is the word β€œapproaches.” We are interested in the behavior of the function near cc, not necessarily at cc itself. This is particularly crucial when dealing with functions that are not defined at a specific point.

Limits form the bedrock of calculus, underpinning the concepts of continuity, derivatives, and integrals. Without a solid grasp of limits, one cannot fully comprehend these more advanced topics. Limits are used extensively in various fields, including physics, engineering, economics, and computer science, to model and analyze real-world phenomena. For instance, in physics, limits are used to define instantaneous velocity and acceleration. In economics, they are used to analyze marginal cost and marginal revenue. Understanding limits is not just about manipulating equations; it's about understanding the fundamental behavior of functions and their applications in various contexts.

Problem Statement: lim⁑xβ†’3x2+xβˆ’12x2βˆ’3x\lim_{x \rightarrow 3} \frac{x^2+x-12}{x^2-3x}

Let's focus on the specific limit we aim to solve: lim⁑xβ†’3x2+xβˆ’12x2βˆ’3x\lim_{x \rightarrow 3} \frac{x^2+x-12}{x^2-3x}. This expression presents an interesting challenge. If we directly substitute x=3x = 3 into the function, we encounter a problem:

32+3βˆ’1232βˆ’3(3)=9+3βˆ’129βˆ’9=00\frac{3^2 + 3 - 12}{3^2 - 3(3)} = \frac{9 + 3 - 12}{9 - 9} = \frac{0}{0}

We obtain 00\frac{0}{0}, which is an indeterminate form. This means that we cannot determine the limit simply by substituting the value. Indeterminate forms signal that we need to employ alternative techniques to evaluate the limit. These techniques often involve algebraic manipulation, such as factoring, simplifying, or rationalizing the expression. The goal is to transform the expression into a form where we can directly substitute the value without encountering the indeterminate form. In this case, the presence of the indeterminate form 00\frac{0}{0} indicates that there might be a common factor in the numerator and the denominator that we can cancel out. This will be the key to solving this limit.

Step-by-Step Solution

To find the limit lim⁑xβ†’3x2+xβˆ’12x2βˆ’3x\lim_{x \rightarrow 3} \frac{x^2+x-12}{x^2-3x}, we will proceed through a series of steps:

Step 1: Factoring the Numerator and Denominator

The first step in evaluating this limit is to factor both the numerator and the denominator. Factoring helps us identify common factors that can be canceled out, which is crucial when dealing with indeterminate forms. The numerator is a quadratic expression, x2+xβˆ’12x^2 + x - 12. We need to find two numbers that multiply to -12 and add up to 1. These numbers are 4 and -3. Therefore, we can factor the numerator as follows:

x2+xβˆ’12=(x+4)(xβˆ’3)x^2 + x - 12 = (x + 4)(x - 3)

Now, let's factor the denominator, x2βˆ’3xx^2 - 3x. This expression has a common factor of xx, so we can factor it as:

x2βˆ’3x=x(xβˆ’3)x^2 - 3x = x(x - 3)

Step 2: Rewriting the Expression

Now that we have factored both the numerator and the denominator, we can rewrite the original expression:

x2+xβˆ’12x2βˆ’3x=(x+4)(xβˆ’3)x(xβˆ’3)\frac{x^2 + x - 12}{x^2 - 3x} = \frac{(x + 4)(x - 3)}{x(x - 3)}

This rewritten form makes it easier to see the common factor that we can cancel out.

Step 3: Simplifying the Expression

We observe that (xβˆ’3)(x - 3) is a common factor in both the numerator and the denominator. As long as xx is not equal to 3, we can cancel this factor out. This is a crucial step because we are considering the limit as xx approaches 3, not when xx equals 3. Therefore, we can simplify the expression:

(x+4)(xβˆ’3)x(xβˆ’3)=x+4x\frac{(x + 4)(x - 3)}{x(x - 3)} = \frac{x + 4}{x}

Step 4: Evaluating the Limit

Now that we have simplified the expression, we can evaluate the limit by substituting x=3x = 3 into the simplified expression:

lim⁑xβ†’3x+4x=3+43=73\lim_{x \rightarrow 3} \frac{x + 4}{x} = \frac{3 + 4}{3} = \frac{7}{3}

Thus, the limit of the function as xx approaches 3 is 73\frac{7}{3}.

Conclusion

In this article, we successfully evaluated the limit lim⁑xβ†’3x2+xβˆ’12x2βˆ’3x\lim_{x \rightarrow 3} \frac{x^2+x-12}{x^2-3x}. We encountered an indeterminate form 00\frac{0}{0} when directly substituting x=3x = 3, which necessitated the use of algebraic manipulation. By factoring both the numerator and the denominator, we identified a common factor (xβˆ’3)(x - 3) that could be canceled out. This simplification allowed us to evaluate the limit by direct substitution, resulting in the final answer of 73\frac{7}{3}.

The process we followed highlights the importance of algebraic techniques in evaluating limits. Factoring, simplifying, and canceling common factors are powerful tools in the calculus toolkit. This example illustrates how a seemingly complex limit problem can be solved with a methodical approach and a solid understanding of algebraic principles. Mastering these techniques is essential for tackling more advanced calculus problems and for applying limits in various fields of science and engineering. Understanding limits is not just a mathematical exercise; it is a foundational skill for anyone working with continuous change and dynamic systems.