Evaluating Limit Of (3-√(x+4))/(x-5) A Comprehensive Guide

$$

Introduction

In the realm of calculus, limits form the bedrock upon which the concepts of continuity, derivatives, and integrals are built. Understanding limits is crucial for grasping the behavior of functions as their input approaches a particular value. This article delves into the evaluation of a specific limit: ${\lim _{x \rightarrow 5} \frac{3-\sqrt{x+4}}{x-5}}$. This limit exemplifies a common scenario in calculus where direct substitution leads to an indeterminate form, specifically 0/0. To navigate this, we will explore algebraic techniques, particularly rationalization, to transform the expression into a form where the limit can be readily computed. This comprehensive guide aims to provide a step-by-step approach, ensuring clarity and a deep understanding of the underlying principles. By mastering such limit evaluations, you will enhance your calculus skills and be better equipped to tackle more complex problems. So, let’s embark on this journey of mathematical exploration and unravel the intricacies of this limit.

Understanding the Limit Concept

The concept of a limit is fundamental to calculus and mathematical analysis. At its core, a limit describes the value that a function approaches as the input (or argument) approaches a certain value. More formally, we say that the limit of a function f(x) as x approaches c is L, denoted as ${\lim_{x \rightarrow c} f(x) = L}$, if the values of f(x) become arbitrarily close to L as x gets arbitrarily close to c, but not necessarily equal to c. This definition highlights a crucial aspect of limits: we are concerned with the behavior of the function near the point c, not necessarily at the point c itself.

Limits are essential for defining continuity. A function f(x) is continuous at a point c if three conditions are met: f(c) is defined, ${\lim_{x \rightarrow c} f(x)}$ exists, and ${\lim_{x \rightarrow c} f(x) = f(c)}$. In simpler terms, a function is continuous if you can draw its graph without lifting your pen from the paper. Limits also play a vital role in defining derivatives, which represent the instantaneous rate of change of a function. The derivative of f(x) at a point x is defined as the limit of the difference quotient: ${f'(x) = \lim_{h \rightarrow 0} \frac{f(x+h) - f(x)}{h}}$. This limit gives us the slope of the tangent line to the graph of f(x) at the point x, providing valuable information about the function's behavior.

When evaluating limits, several techniques can be employed. Direct substitution is the simplest method, where we substitute the value that x approaches into the function. However, this method often fails when we encounter indeterminate forms such as 0/0, ∞/∞, or 0 * ∞. In such cases, algebraic manipulation, rationalization, factoring, or L'Hôpital's Rule may be necessary to transform the expression into a form where the limit can be evaluated. Understanding the concept of limits and mastering these evaluation techniques are crucial for success in calculus and related fields. They allow us to analyze the behavior of functions, understand continuity and differentiability, and solve a wide range of problems in mathematics, physics, engineering, and economics. The limit we are about to explore, ${\lim _{x \rightarrow 5} \frac{3-\sqrt{x+4}}{x-5}}$, falls into the category where direct substitution results in an indeterminate form, necessitating the use of algebraic manipulation to find the solution.

Identifying the Indeterminate Form

Before diving into the solution, it's crucial to understand why the given limit, ${\lim _{x \rightarrow 5} \frac{3-\sqrt{x+4}}{x-5}}$, requires special attention. The first step in evaluating any limit is to attempt direct substitution. This involves simply replacing the variable x with the value it approaches, in this case, 5. If we substitute x = 5 directly into the expression, we get:

${\frac{3-\sqrt{5+4}}{5-5} = \frac{3-\sqrt{9}}{0} = \frac{3-3}{0} = \frac{0}{0}}$

The result, 0/0, is an indeterminate form. This means that the value of the limit cannot be determined directly from the substitution. The 0/0 form indicates that there is a potential for cancellation or simplification within the expression. It doesn't necessarily mean that the limit doesn't exist; it simply means that further analysis is required to determine its value. Other common indeterminate forms include ∞/∞, 0 * ∞, ∞ - ∞, 1^∞, 0^0, and ∞^0. Each of these forms requires a different approach to evaluate the limit.

The presence of an indeterminate form signals the need for algebraic manipulation or other techniques to rewrite the expression in a form where the limit can be evaluated. In the case of ${\lim _{x \rightarrow 5} \frac{3-\sqrt{x+4}}{x-5}}$, the presence of the square root in the numerator suggests that rationalization might be a suitable technique. Rationalization involves multiplying the numerator and denominator by the conjugate of the expression containing the square root. This process aims to eliminate the square root from either the numerator or the denominator, potentially simplifying the expression and allowing us to evaluate the limit. Recognizing the indeterminate form is a critical first step in limit evaluation, as it guides the choice of appropriate techniques to solve the problem. In the next section, we will explore the rationalization technique in detail and apply it to the given limit.

Applying Rationalization

Since direct substitution resulted in the indeterminate form 0/0, we need to employ a technique to manipulate the expression ${\frac{3-\sqrt{x+4}}{x-5}}$ and make the limit evaluation possible. The presence of the square root term, ${\sqrt{x+4}}$, in the numerator suggests that rationalization is the appropriate strategy. Rationalization involves multiplying both the numerator and the denominator of the fraction by the conjugate of the expression containing the square root. The conjugate is formed by changing the sign between the terms. In this case, the conjugate of ${3-\sqrt{x+4}}$ is ${3+\sqrt{x+4}}$.

Let's perform the rationalization:

${\lim _{x \rightarrow 5} \frac{3-\sqrt{x+4}}{x-5} = \lim _{x \rightarrow 5} \frac{3-\sqrt{x+4}}{x-5} \cdot \frac{3+\sqrt{x+4}}{3+\sqrt{x+4}}}$

Multiplying the numerators, we use the difference of squares formula, (a - b)(a + b) = a^2 - b^2:

${= \lim _{x \rightarrow 5} \frac{(3)^2 - (\sqrt{x+4})^2}{(x-5)(3+\sqrt{x+4})}}$

${= \lim _{x \rightarrow 5} \frac{9 - (x+4)}{(x-5)(3+\sqrt{x+4})}}$

Simplifying the numerator:

${= \lim _{x \rightarrow 5} \frac{9 - x - 4}{(x-5)(3+\sqrt{x+4})}}$

${= \lim _{x \rightarrow 5} \frac{5 - x}{(x-5)(3+\sqrt{x+4})}}$

Notice that (5 - x) is the negative of (x - 5). We can factor out a -1 from the numerator:

${= \lim _{x \rightarrow 5} \frac{-1(x - 5)}{(x-5)(3+\sqrt{x+4})}}$

Now, we can cancel the common factor of (x - 5) from the numerator and the denominator, provided that x ≠ 5. This is valid because we are considering the limit as x approaches 5, not the value of the function at x = 5:

${= \lim _{x \rightarrow 5} \frac{-1}{3+\sqrt{x+4}}}$

By applying rationalization, we have successfully transformed the original expression into a simpler form where direct substitution is now possible. In the next section, we will evaluate the limit using direct substitution on this simplified expression.

Evaluating the Simplified Limit

After applying rationalization, we have simplified the original limit expression to:

${\lim _{x \rightarrow 5} \frac{-1}{3+\sqrt{x+4}}}$

Now, we can attempt direct substitution again. Substituting x = 5 into the simplified expression, we get:

${= \frac{-1}{3+\sqrt{5+4}}}$

${= \frac{-1}{3+\sqrt{9}}}$

${= \frac{-1}{3+3}}$

${= \frac{-1}{6}}$

Therefore, the limit is:

${\lim _{x \rightarrow 5} \frac{3-\sqrt{x+4}}{x-5} = -\frac{1}{6}}$

By successfully rationalizing the original expression and then applying direct substitution, we have found the value of the limit. This demonstrates the power of algebraic manipulation in evaluating limits that initially present indeterminate forms. The key was to recognize the indeterminate form 0/0, apply the appropriate technique (rationalization), and simplify the expression to a point where direct substitution yielded the solution.

This result provides valuable information about the behavior of the function ${f(x) = \frac{3-\sqrt{x+4}}{x-5}}$ as x approaches 5. The function approaches -1/6, even though the function is not defined at x = 5 (due to the original denominator becoming zero). This highlights the essence of a limit: it describes the function's behavior near a point, not necessarily at the point itself.

In conclusion, evaluating limits often requires a combination of techniques, including recognizing indeterminate forms, applying algebraic manipulations, and finally, direct substitution. The rationalization technique, as demonstrated in this example, is a powerful tool for dealing with expressions involving square roots. Mastering these techniques is crucial for a deeper understanding of calculus and its applications.

Alternative Methods and Further Insights

While rationalization is an effective method for evaluating the limit ${\lim _{x \rightarrow 5} \frac{3-\sqrt{x+4}}{x-5}}$, it is not the only approach. Understanding alternative methods can provide a more comprehensive understanding of limit evaluation and offer different perspectives on the problem. One such alternative is L'Hôpital's Rule.

L'Hôpital's Rule states that if the limit of ${\frac{f(x)}{g(x)}}$ as x approaches c results in an indeterminate form (0/0 or ∞/∞), and if f and g are differentiable functions, then:

${\lim_{x \rightarrow c} \frac{f(x)}{g(x)} = \lim_{x \rightarrow c} \frac{f'(x)}{g'(x)}}$

provided the limit on the right-hand side exists. In simpler terms, if you have an indeterminate form, you can take the derivative of the numerator and the derivative of the denominator separately and then try evaluating the limit again.

Let's apply L'Hôpital's Rule to our limit. First, we identify f(x) and g(x):

• f(x) = 3 - ${\sqrt{x+4}}$
• g(x) = x - 5

Now, we find the derivatives f'(x) and g'(x):

• f'(x) = -${\frac{1}{2\sqrt{x+4}}}$ (using the chain rule)
• g'(x) = 1

Applying L'Hôpital's Rule:

${\lim _{x \rightarrow 5} \frac{3-\sqrt{x+4}}{x-5} = \lim _{x \rightarrow 5} \frac{-\frac{1}{2\sqrt{x+4}}}{1}}$

Now, we can use direct substitution:

${= \frac{-\frac{1}{2\sqrt{5+4}}}{1} = \frac{-\frac{1}{2\sqrt{9}}}{1} = \frac{-\frac{1}{2 \cdot 3}}{1} = -\frac{1}{6}}$

As we can see, L'Hôpital's Rule provides the same result, -1/6, as the rationalization method. This illustrates the versatility of calculus techniques in solving limit problems.

Furthermore, it's important to recognize the graphical interpretation of limits. The limit ${\lim _{x \rightarrow 5} \frac{3-\sqrt{x+4}}{x-5}}$ represents the slope of the tangent line to the function ${h(x) = 3-\sqrt{x+4}}$ at the point where x = 5. By understanding this connection, we can visualize the limit and gain a deeper understanding of its meaning.

In addition to these methods, series expansions can also be used to evaluate limits, especially for more complex functions. However, for this particular limit, rationalization and L'Hôpital's Rule are the most straightforward approaches. Exploring these alternative methods not only provides additional tools for limit evaluation but also reinforces the interconnectedness of calculus concepts.

Conclusion

In this comprehensive guide, we have explored the evaluation of the limit ${\lim _{x \rightarrow 5} \frac{3-\sqrt{x+4}}{x-5}}$. We began by understanding the fundamental concept of limits and the importance of identifying indeterminate forms. Direct substitution led us to the indeterminate form 0/0, which necessitated the use of algebraic manipulation.

We successfully applied the rationalization technique, multiplying the numerator and denominator by the conjugate of the expression containing the square root. This allowed us to simplify the expression and eliminate the indeterminate form. After simplification, direct substitution yielded the limit value of -1/6.

Furthermore, we explored an alternative method, L'Hôpital's Rule, which provided an independent confirmation of our result. This highlighted the versatility of calculus techniques and the importance of understanding multiple approaches to problem-solving. We also discussed the graphical interpretation of limits, connecting the concept to the slope of a tangent line.

Mastering limit evaluation techniques is crucial for success in calculus and related fields. The ability to recognize indeterminate forms, apply appropriate algebraic manipulations, and utilize tools like L'Hôpital's Rule empowers you to tackle a wide range of problems. The limit ${\lim _{x \rightarrow 5} \frac{3-\sqrt{x+4}}{x-5}}$ serves as a valuable example, illustrating the power and elegance of calculus in describing the behavior of functions. By continuing to practice and explore different types of limits, you will strengthen your understanding and develop the skills necessary to excel in your mathematical journey. Remember, the journey of mathematical exploration is a continuous one, and each limit problem solved is a step forward in deepening your understanding of the subject.