Evaluate Limit Of (x^2 - 16) / (1 - √(x - 3)) As X Approaches 4

This article delves into the process of evaluating the limit of the expression ${ \frac{x^2 - 16}{1 - \sqrt{x - 3}} }$ as x approaches 4. This is a classic calculus problem that requires careful manipulation and application of limit laws. We'll explore the techniques needed to solve this, including factoring, rationalizing the denominator, and direct substitution. Understanding limits is fundamental in calculus, as they form the basis for concepts such as continuity, derivatives, and integrals. This particular problem showcases how algebraic manipulation can simplify complex expressions and allow us to find limits that are not immediately obvious.

Understanding Limits

Before we dive into the solution, let's briefly discuss the concept of limits. In calculus, a limit describes the value that a function approaches as the input (or independent variable) approaches a certain value. More formally, the limit of a function f(x) as x approaches c is the value L that f(x) gets arbitrarily close to as x gets arbitrarily close to c, but not necessarily equal to c. This is denoted as:

${ \lim_{x \to c} f(x) = L }$

Limits are crucial because they allow us to analyze the behavior of functions near points where the function might not be defined, or where direct substitution leads to indeterminate forms like 0/0. In such cases, we need to employ algebraic techniques or other methods to evaluate the limit. The concept of limits is fundamental to calculus, underpinning derivatives, integrals, and continuity. Understanding limits allows us to analyze function behavior near specific points. This involves examining what happens to the output of a function, f(x), as the input, x, gets closer and closer to a particular value, often denoted as c. However, the limit focuses on the value the function approaches, not necessarily the value of the function at c. This distinction is crucial because functions may not be defined at c, or their value at c might differ from the limit. The formal definition of a limit refines this idea by stating that for any small positive number (often denoted as epsilon, ε), there exists another small positive number (delta, δ) such that if x is within δ of c, then f(x) is within ε of the limit L. This rigorous definition helps us handle potentially problematic cases, such as functions with discontinuities or those approaching infinity. The limit of a function as x approaches c is written as lim (x→c) f(x) = L, where L is the value the function approaches. If directly substituting c into the function results in an indeterminate form like 0/0, then algebraic manipulation or other techniques are required to find the limit. This could involve factoring, rationalizing, or applying L'Hôpital's Rule, each designed to transform the expression into a form where the limit can be evaluated more directly. Understanding these techniques is essential for solving a wide range of calculus problems and for grasping more advanced concepts built upon limits. Without limits, concepts like instantaneous rate of change (derivatives) and the area under a curve (integrals) wouldn't be defined. Thus, mastering limits is a foundational step in calculus education, and its importance cannot be overstated. This is why problems like evaluating the limit of (x^2 - 16) / (1 - √(x - 3)) as x approaches 4 are so valuable; they reinforce the fundamental techniques for working with limits. The process involves recognizing the indeterminate form, applying algebraic manipulation to simplify the expression, and then finding the limit through direct substitution or other appropriate methods.

Problem Setup

We are tasked with evaluating the following limit:

${ \lim_{x \to 4} \frac{x^2 - 16}{1 - \sqrt{x - 3}} }$

If we attempt to directly substitute x = 4 into the expression, we get:

${ \frac{4^2 - 16}{1 - \sqrt{4 - 3}} = \frac{16 - 16}{1 - \sqrt{1}} = \frac{0}{1 - 1} = \frac{0}{0} }$

This is an indeterminate form, which means we cannot directly evaluate the limit by substitution. We need to manipulate the expression algebraically to find an equivalent form where the limit can be determined. The initial evaluation of the limit, by direct substitution, results in an indeterminate form of 0/0. This tells us we can't simply plug in x = 4 to find the limit. Instead, this outcome signals the need for algebraic manipulation to simplify the expression. Indeterminate forms arise when both the numerator and denominator of a fraction approach zero (or infinity) simultaneously. This doesn't mean the limit doesn't exist, but rather that the initial form of the expression is obscuring the limit's true value. The most common techniques for dealing with such indeterminate forms involve factoring, rationalizing, and, in more advanced cases, L'Hôpital's Rule. In this specific problem, we see a difference of squares in the numerator (x^2 - 16) and a square root in the denominator, which suggests that factoring and rationalization are the appropriate strategies. The presence of the square root term, 1 - √(x - 3), particularly points towards rationalization as the next step. Rationalizing the denominator involves multiplying both the numerator and denominator by the conjugate of the denominator. This process eliminates the square root from the denominator, potentially simplifying the expression and allowing us to evaluate the limit. The key to solving this problem lies in identifying and addressing the indeterminate form. Understanding that 0/0 is not a defined value but rather an indication of further work needed is crucial. From there, the path to the solution involves choosing the right algebraic techniques to transform the expression into a more manageable form. Each algebraic step is taken with the aim of canceling out terms that are causing the indeterminate form, eventually leading to an expression where direct substitution is possible. This process highlights a fundamental principle in calculus: limits are about the behavior of a function near a point, not necessarily at the point, and algebraic manipulation is often required to reveal this behavior. The initial setup, identifying the indeterminate form, and planning the algebraic approach are critical first steps in solving limit problems. Without a clear understanding of these foundational elements, the subsequent steps, such as the rationalization of the denominator, may not be apparent or effective.

Rationalizing the Denominator

To eliminate the square root in the denominator, we multiply both the numerator and the denominator by the conjugate of the denominator, which is (1 + √(x - 3)):

${ \lim_{x \to 4} \frac{x^2 - 16}{1 - \sqrt{x - 3}} \cdot \frac{1 + \sqrt{x - 3}}{1 + \sqrt{x - 3}} }$

This gives us:

${ \lim_{x \to 4} \frac{(x^2 - 16)(1 + \sqrt{x - 3})}{1 - (x - 3)} }$

Simplifying the denominator:

${ \lim_{x \to 4} \frac{(x^2 - 16)(1 + \sqrt{x - 3})}{4 - x} }$

Rationalizing the denominator is a crucial technique when dealing with limits involving square roots. The process involves multiplying both the numerator and denominator of the expression by the conjugate of the denominator. The conjugate is formed by changing the sign between the terms in the denominator. In our case, the conjugate of (1 - √(x - 3)) is (1 + √(x - 3)). Multiplying by the conjugate is strategically done because it leverages the difference of squares identity: (a - b)(a + b) = a^2 - b^2. This eliminates the square root from the denominator, transforming the expression into a form that is often easier to manipulate. The act of multiplying by the conjugate is equivalent to multiplying by 1, so the value of the expression remains unchanged. However, the algebraic form is significantly altered, which is precisely what we need to address the indeterminate form. After multiplying by the conjugate, the expression becomes [(x^2 - 16)(1 + √(x - 3))] / [1 - (x - 3)]. The next step is to simplify the denominator. Distributing the negative sign in the denominator, we get 1 - x + 3, which simplifies to 4 - x. This simplification is essential because it sets the stage for further algebraic manipulation, specifically factoring and cancellation. The rationalization step is a key transformation that allows us to bypass the indeterminate form. It highlights the importance of algebraic manipulation in calculus, where the goal is often to rewrite an expression in a way that makes the limit easier to evaluate. While it may seem like a simple algebraic trick, it's a powerful tool for solving a wide range of limit problems, especially those involving radicals. This step also exemplifies a broader theme in problem-solving: identifying and addressing the obstacles that prevent us from reaching a solution. In the case of limits, the indeterminate form is the obstacle, and rationalization is the tool to overcome it. Without rationalizing, the limit would remain difficult to evaluate, and the subsequent steps of factoring and canceling would not be possible. Thus, understanding and applying rationalization is a core skill in calculus and an essential part of evaluating limits.

Factoring and Simplifying

Now, we can factor the numerator. x² - 16 is a difference of squares, which factors as (x - 4)(x + 4):

${ \lim_{x \to 4} \frac{(x - 4)(x + 4)(1 + \sqrt{x - 3})}{4 - x} }$

Notice that (4 - x) is the negative of (x - 4). We can rewrite the expression as:

${ \lim_{x \to 4} \frac{-(4 - x)(x + 4)(1 + \sqrt{x - 3})}{4 - x} }$

Now, we can cancel the (4 - x) terms:

${ \lim_{x \to 4} -(x + 4)(1 + \sqrt{x - 3}) }$

Factoring and simplifying is a vital step in evaluating limits, particularly after rationalizing or when dealing with polynomial expressions. Factoring involves breaking down a complex expression into simpler components, often products of binomials or other polynomials. In this specific problem, the numerator contains the expression x^2 - 16, which is a classic example of a difference of squares. Recognizing this pattern is crucial because it allows us to factor it into (x - 4)(x + 4). This factorization is significant because it reveals a term, (x - 4), that can potentially cancel with a similar term in the denominator. The denominator, after rationalization, is (4 - x). It's important to notice that (4 - x) is the negative of (x - 4). This observation is key to simplifying the expression further. To facilitate cancellation, we rewrite the expression as - (x - 4) in the denominator, or equivalently, we factor out a -1 from either the numerator or the denominator. This allows us to clearly see the common factor of (x - 4) in both the numerator and the denominator. The cancellation of this common factor is a critical step in resolving the indeterminate form. By removing the factor that causes both the numerator and denominator to approach zero, we pave the way for direct substitution. The ability to recognize and apply factoring techniques is a fundamental skill in algebra and calculus. It’s essential for simplifying expressions, solving equations, and, as we see here, evaluating limits. This step underscores the importance of algebraic manipulation in calculus problem-solving. Factoring, combined with cancellation, transforms the expression into a form where the limit can be easily determined. Without factoring and simplifying, the indeterminate form would persist, and the limit would remain elusive. The elegance of this step lies in its ability to reveal the underlying structure of the expression, making the limit evaluation straightforward. The combination of rationalizing and factoring is a powerful approach for a wide range of limit problems. It demonstrates how different algebraic techniques can be combined to overcome obstacles and arrive at a solution. This strategic combination of tools is a hallmark of effective problem-solving in calculus.

Direct Substitution

Now, we can directly substitute x = 4 into the simplified expression:

${ -(4 + 4)(1 + \sqrt{4 - 3}) = -(8)(1 + \sqrt{1}) = -(8)(1 + 1) = -(8)(2) = -16 }$

Therefore, the limit is -16.

Direct substitution is the final step in evaluating many limits, but it can only be applied after the expression has been simplified to a point where it no longer results in an indeterminate form. After the algebraic manipulations of rationalization and factoring, we arrive at the simplified expression: - (x + 4)(1 + √(x - 3)). This expression is now suitable for direct substitution because plugging in x = 4 will not lead to division by zero or any other indeterminate form. Direct substitution involves replacing every instance of the variable x in the expression with the value that x is approaching, in this case, 4. So, we substitute x = 4 into the simplified expression: - (4 + 4)(1 + √(4 - 3)). The next step is to evaluate the expression. First, we simplify the terms inside the parentheses: - (8)(1 + √1). Then, we evaluate the square root: - (8)(1 + 1). Next, we add the terms inside the second set of parentheses: - (8)(2). Finally, we multiply: -16. The result of direct substitution is the value of the limit. In this case, the limit as x approaches 4 of the original expression is -16. This step highlights the importance of the preceding algebraic manipulations. Without rationalizing and factoring, direct substitution would have resulted in the indeterminate form 0/0, and we would not have been able to find the limit. The ability to perform direct substitution is a clear sign that the expression has been successfully simplified. It represents the culmination of the problem-solving process, where all the necessary algebraic techniques have been applied to reveal the limit's value. This step also underscores the definition of a limit. The limit is the value that the function approaches as x approaches a particular value, and direct substitution, when applicable, allows us to find that value. The success of direct substitution depends on the quality of the preceding steps. If the expression is not simplified correctly, direct substitution may lead to an incorrect answer or an indeterminate form, indicating that further manipulation is required. Therefore, each step in the process, from rationalizing to factoring to simplifying, is crucial for arriving at the correct solution. In summary, direct substitution is the final act in the limit evaluation process, providing the answer after the expression has been carefully prepared through algebraic manipulation.

Conclusion

By rationalizing the denominator, factoring, and simplifying, we were able to evaluate the limit of the given expression as x approaches 4. The correct answer is -16. This problem demonstrates the importance of algebraic manipulation in evaluating limits, especially when dealing with indeterminate forms. In conclusion, the evaluation of the limit lim (x→4) [(x^2 - 16) / (1 - √(x - 3))] required a series of strategic algebraic manipulations. We began by recognizing that direct substitution led to an indeterminate form, 0/0, indicating the need for further simplification. The first key step was rationalizing the denominator by multiplying both the numerator and denominator by the conjugate of the denominator, (1 + √(x - 3)). This eliminated the square root in the denominator, transforming the expression into a more manageable form. Next, we factored the numerator, recognizing the difference of squares pattern in (x^2 - 16). This factorization allowed us to identify a common factor with the simplified denominator. We then simplified the expression by canceling the common factor, which was crucial for resolving the indeterminate form. Finally, with the expression simplified, we applied direct substitution, plugging in x = 4 to find the limit. The result of this process was -16, which is the value the function approaches as x gets closer to 4. The solution highlights several important concepts in calculus and problem-solving. It demonstrates the power of algebraic manipulation in transforming complex expressions into simpler forms. It also underscores the significance of recognizing patterns, such as the difference of squares, and applying appropriate techniques, such as rationalization and factoring. Furthermore, the problem illustrates the fundamental idea of a limit: we're interested in the behavior of a function near a point, not necessarily at the point. The initial indeterminate form prevented us from directly evaluating the function at x = 4, but by using limits, we were able to determine the value the function approaches as x gets arbitrarily close to 4. This type of problem is a cornerstone of calculus education because it reinforces essential skills and concepts. The combination of algebraic techniques and the understanding of limits is crucial for tackling more advanced topics in calculus, such as derivatives and integrals. The ability to solve problems like this is a testament to a solid foundation in algebra and a growing understanding of the core principles of calculus.

Final Answer: (A) -16