Evaluating Limits A Comprehensive Guide To Lim X→-5 (x^2-2x-35)/(7x^2+34x-5)

Introduction to Limits in Calculus

In the realm of calculus, limits stand as a fundamental concept, serving as the bedrock upon which more advanced topics such as derivatives and integrals are built. Understanding limits is crucial for grasping the behavior of functions as their input approaches a specific value. This article delves into the evaluation of a particular limit problem, offering a step-by-step solution and insightful explanations to enhance comprehension. The limit we aim to evaluate is:

$$\lim _{x \rightarrow-5} \frac{x^2-2 x-35}{7 x^2+34 x-5}$$

This expression represents the value the function ${ f(x) = \frac{x^2-2 x-35}{7 x^2+34 x-5} }$ approaches as ${ x }$ gets arbitrarily close to -5. The journey to find this limit involves algebraic manipulation, factoring, and ultimately, direct substitution. This process not only yields the answer but also reinforces key concepts in calculus, making it a valuable exercise for students and enthusiasts alike.

Understanding the Limit Problem

Before diving into the solution, let's dissect the problem at hand. We are tasked with evaluating the limit of a rational function as ${ x }$ approaches -5. The function is given by ${ \frac{x^2-2 x-35}{7 x^2+34 x-5} }$. At first glance, one might be tempted to directly substitute ${ x = -5 }$ into the function. However, doing so results in a ${ \frac{0}{0} }$ form, which is an indeterminate form. This means we cannot determine the limit directly and need to employ algebraic techniques to simplify the expression. The indeterminate form signals the presence of a common factor in both the numerator and the denominator that becomes zero when ${ x = -5 }$. Identifying and canceling this factor is key to finding the limit. This situation underscores the importance of understanding algebraic manipulation in calculus, particularly when dealing with limits. Factoring quadratic expressions, in this case, is the critical step towards unlocking the solution. By carefully factoring both the numerator and the denominator, we can reveal the common factor and simplify the expression, paving the way for direct substitution. This process not only solves the problem but also highlights the elegance and power of algebraic techniques in resolving calculus problems.

Step-by-Step Solution: Evaluating the Limit

To effectively evaluate the limit ${ \lim _{x \rightarrow-5} \frac{x^2-2 x-35}{7 x^2+34 x-5} }$, we will proceed through a series of logical steps, primarily focusing on algebraic manipulation to resolve the indeterminate form. The first key step in evaluating this limit involves factoring both the numerator and the denominator of the rational function. This is crucial because direct substitution of ${ x = -5 }$ leads to the indeterminate form ${ \frac{0}{0} }$, which does not provide a clear answer about the limit's value.

Step 1: Factoring the Numerator

The numerator is a quadratic expression: ${ x^2 - 2x - 35 }$. To factor it, we look for two numbers that multiply to -35 and add to -2. These numbers are -7 and 5. Thus, the numerator can be factored as:

$$x^2 - 2x - 35 = (x - 7)(x + 5)$$

The ability to factor quadratic expressions accurately is fundamental in calculus, particularly when dealing with limits and rational functions. This step transforms the numerator into a product of two binomials, which is essential for identifying any common factors with the denominator.

Step 2: Factoring the Denominator

The denominator is another quadratic expression: ${ 7x^2 + 34x - 5 }$. Factoring this requires a bit more effort since the leading coefficient is not 1. We look for two numbers that multiply to ${ 7 \times -5 = -35 }$ and add to 34. These numbers are 35 and -1. We can rewrite the middle term using these numbers and then factor by grouping:

$$7x^2 + 34x - 5 = 7x^2 + 35x - x - 5$$

Now, factor by grouping:

$$= 7x(x + 5) - 1(x + 5)$$

$$= (7x - 1)(x + 5)$$

Factoring the denominator is a critical step as it reveals a common factor with the factored numerator, which is essential for simplifying the expression and resolving the indeterminate form. The technique of factoring by grouping is particularly useful when the leading coefficient is not 1, making it a valuable tool in algebraic manipulation.

Step 3: Simplifying the Rational Function

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

$$\frac{x^2-2 x-35}{7 x^2+34 x-5} = \frac{(x - 7)(x + 5)}{(7x - 1)(x + 5)}$$

We notice that ${ (x + 5) }$ is a common factor in both the numerator and the denominator. We can cancel this factor, provided that ${ x \neq -5 }$ (since division by zero is undefined). This simplification is a key step in evaluating limits, as it removes the indeterminate form and allows for direct substitution.

$$\frac{(x - 7)(x + 5)}{(7x - 1)(x + 5)} = \frac{x - 7}{7x - 1}$$

This simplification transforms the original rational function into a more manageable form, making the evaluation of the limit significantly easier. The cancellation of the common factor highlights the importance of algebraic manipulation in resolving limit problems.

Step 4: Evaluating the Limit by Direct Substitution

After simplifying the rational function, we can now evaluate the limit by direct substitution. We substitute ${ x = -5 }$ into the simplified expression:

$$\lim _{x \rightarrow-5} \frac{x - 7}{7x - 1} = \frac{-5 - 7}{7(-5) - 1}$$

$$= \frac{-12}{-35 - 1}$$

$$= \frac{-12}{-36}$$

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

Direct substitution, after simplification, is a powerful technique for evaluating limits. It transforms the problem into a straightforward arithmetic calculation, providing a clear and precise answer. This step demonstrates the culmination of the previous algebraic manipulations, leading to the final result.

Therefore, the limit of the given function as ${ x }$ approaches -5 is ${ \frac{1}{3} }$.

$$\lim _{x \rightarrow-5} \frac{x^2-2 x-35}{7 x^2+34 x-5} = \frac{1}{3}$$

This comprehensive step-by-step solution illustrates the process of evaluating limits, emphasizing the importance of factoring, simplification, and direct substitution. The final answer, ${ \frac{1}{3} }$, represents the value the function approaches as ${ x }$ gets arbitrarily close to -5.

Common Mistakes to Avoid When Evaluating Limits

Evaluating limits can sometimes be tricky, and there are several common mistakes that students often make. Being aware of these pitfalls can help in avoiding them and ensuring accurate solutions. One frequent error is attempting to directly substitute the value into the function without first checking for indeterminate forms. As seen in our example, directly substituting ${ x = -5 }$ into the original function results in ${ \frac{0}{0} }$, which is an indeterminate form. This means the limit cannot be determined by direct substitution alone, and further algebraic manipulation is required.

Another common mistake is incorrect factoring. Factoring quadratic expressions is a crucial step in many limit problems, and errors in factoring can lead to incorrect simplifications and, ultimately, the wrong answer. It's essential to double-check the factored expressions to ensure they are correct. For instance, in our problem, accurately factoring both the numerator and the denominator was key to identifying the common factor and simplifying the expression.

Canceling terms incorrectly is another pitfall. While canceling common factors is a valid technique for simplifying rational functions, it must be done carefully. Factors can only be canceled if they are multiplied by the entire numerator and the entire denominator. Incorrectly canceling terms that are added or subtracted can lead to significant errors. In our example, canceling the ${ (x + 5) }$ factor was valid because it was a factor of both the numerator and the denominator.

Forgetting to simplify the expression fully before evaluating the limit is also a common mistake. Simplifying the expression as much as possible before substituting the value can make the evaluation process much easier and reduce the chances of error. In our case, simplifying the rational function by canceling the common factor made the direct substitution straightforward.

Finally, misinterpreting the meaning of a limit is a conceptual mistake that can lead to misunderstandings. A limit represents the value a function approaches as its input approaches a certain value, but it doesn't necessarily equal the function's value at that point. This distinction is particularly important when dealing with functions that have discontinuities or are undefined at certain points. By understanding the concept of limits and avoiding these common mistakes, students can confidently tackle a wide range of limit problems and develop a solid foundation in calculus.

Conclusion: Mastering Limit Evaluations

In conclusion, evaluating limits is a fundamental skill in calculus, and mastering it requires a combination of algebraic techniques and conceptual understanding. The step-by-step solution presented in this article demonstrates a systematic approach to solving limit problems, emphasizing the importance of factoring, simplification, and direct substitution. By carefully factoring the numerator and the denominator, we were able to identify and cancel the common factor, transforming the original expression into a simpler form. This allowed us to evaluate the limit by direct substitution, leading to the answer ${ \frac{1}{3} }$.

Throughout this process, we highlighted common mistakes to avoid, such as attempting direct substitution without checking for indeterminate forms, incorrect factoring, and misinterpreting the meaning of a limit. By being aware of these pitfalls, students can improve their accuracy and confidence in solving limit problems. The ability to evaluate limits is not only crucial for calculus but also serves as a foundation for more advanced topics in mathematics and related fields. Limits provide a way to analyze the behavior of functions near specific points, which is essential for understanding continuity, derivatives, and integrals. Mastering limits opens doors to a deeper understanding of mathematical concepts and their applications in the real world.

This article aimed to provide a comprehensive guide to evaluating the limit ${ \lim _{x \rightarrow-5} \frac{x^2-2 x-35}{7 x^2+34 x-5} }$, offering a clear solution and valuable insights. By understanding the techniques and concepts discussed, readers can enhance their problem-solving skills and gain a solid foundation in calculus. The journey through this limit problem underscores the beauty and elegance of mathematics, where careful analysis and algebraic manipulation lead to precise and meaningful results.