Solving Limits A Step-by-Step Guide

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Introduction

In the fascinating realm of calculus, limits serve as a fundamental building block. They form the basis for understanding continuity, derivatives, and integrals. Mastering the techniques for evaluating limits is crucial for anyone venturing into advanced mathematical concepts. This article delves into the intricacies of calculating two specific limits, providing a step-by-step guide and illuminating the underlying principles. We will explore the evaluation of $\lim _{x \rightarrow 1}\left(\frac{\sin (x-1)}{x^2+x-2}\right)$ and $\lim _{x \rightarrow \infty} x \sin \left(\frac{1}{x}\right)$, offering insights that empower you to tackle similar problems with confidence.

1.1.1 Evaluating $\lim _{x \rightarrow 1}\left(\frac{\sin (x-1)}{x^2+x-2}\right)$

To determine the limit $\lim _{x \rightarrow 1}\left(\frac{\sin (x-1)}{x^2+x-2}\right)$, a direct substitution of x=1x = 1 into the expression results in an indeterminate form of 00\frac{0}{0}. This necessitates the use of algebraic manipulation and trigonometric identities to simplify the expression and evaluate the limit. Let's embark on a step-by-step journey to unravel this limit.

Step 1: Factor the denominator

The denominator x2+x−2x^2 + x - 2 can be factored into (x−1)(x+2)(x - 1)(x + 2). This factorization is a crucial step in simplifying the expression and revealing the underlying structure. Factoring allows us to identify common factors that can be canceled, paving the way for a more manageable form. The ability to factor quadratic expressions is a fundamental skill in algebra, and it plays a vital role in solving various mathematical problems.

Step 2: Rewrite the expression

Substituting the factored form of the denominator, the expression becomes:

sin⁡(x−1)(x−1)(x+2)\frac{\sin (x-1)}{(x-1)(x+2)}

This rewritten expression highlights the presence of the term (x−1)(x - 1) in both the numerator and the denominator, which is a key observation for further simplification. Recognizing common factors is a powerful technique in limit evaluation, as it often leads to the cancellation of terms that cause indeterminacy.

Step 3: Apply the special trigonometric limit

A fundamental trigonometric limit states that $\lim_{u \rightarrow 0} \frac{\sin(u)}{u} = 1$. This limit is a cornerstone of calculus and is essential for evaluating many trigonometric limits. To apply this limit, we need to manipulate the expression to match this form. In our case, we can identify uu as (x−1)(x - 1). As xx approaches 1, (x−1)(x - 1) approaches 0, making this limit applicable.

Step 4: Manipulate the expression to match the special limit form

We can rewrite the expression as:

sin⁡(x−1)x−1⋅1x+2\frac{\sin (x-1)}{x-1} \cdot \frac{1}{x+2}

This separation allows us to isolate the term that corresponds to the special trigonometric limit. By strategically manipulating the expression, we have created a form where we can directly apply the known limit. This technique of separating expressions into manageable parts is a common strategy in problem-solving.

Step 5: Evaluate the limit

Now, we can apply the limit as xx approaches 1:

lim⁡x→1(sin⁡(x−1)x−1⋅1x+2)=lim⁡x→1sin⁡(x−1)x−1⋅lim⁡x→11x+2\lim_{x \rightarrow 1} \left(\frac{\sin (x-1)}{x-1} \cdot \frac{1}{x+2}\right) = \lim_{x \rightarrow 1} \frac{\sin (x-1)}{x-1} \cdot \lim_{x \rightarrow 1} \frac{1}{x+2}

Using the special trigonometric limit and direct substitution, we get:

1⋅11+2=131 \cdot \frac{1}{1+2} = \frac{1}{3}

Therefore, the limit $\lim _{x \rightarrow 1}\left(\frac{\sin (x-1)}{x^2+x-2}\right) = \frac{1}{3}$. This result demonstrates the power of combining algebraic manipulation and trigonometric identities to evaluate limits.

1.1.2 Evaluating $\lim _{x \rightarrow \infty} x \sin \left(\frac{1}{x}\right)$

Next, let's tackle the limit $\lim _{x \rightarrow \infty} x \sin \left(\frac{1}{x}\right)$. This limit involves a product of a term approaching infinity and a trigonometric function with an argument approaching zero. This type of limit often requires a clever substitution to transform it into a more manageable form. We will explore the steps involved in evaluating this limit, highlighting the techniques used to overcome the challenges it presents.

Step 1: Introduce a substitution

Let u=1xu = \frac{1}{x}. As xx approaches infinity, uu approaches 0. This substitution is a key step in transforming the limit into a form where we can apply known limit properties. The ability to choose appropriate substitutions is a crucial skill in calculus, as it often simplifies complex expressions and reveals hidden structures.

Step 2: Rewrite the limit in terms of u

Substituting u=1xu = \frac{1}{x}, we have x=1ux = \frac{1}{u}, and the limit becomes:

lim⁡u→01usin⁡(u)=lim⁡u→0sin⁡(u)u\lim_{u \rightarrow 0} \frac{1}{u} \sin(u) = \lim_{u \rightarrow 0} \frac{\sin(u)}{u}

This transformation elegantly converts the original limit into a familiar form. By making the appropriate substitution, we have simplified the problem and made it amenable to a direct application of the special trigonometric limit. This highlights the importance of recognizing patterns and using substitutions to simplify expressions.

Step 3: Apply the special trigonometric limit

As we saw in the previous example, the special trigonometric limit states that $\lim_{u \rightarrow 0} \frac{\sin(u)}{u} = 1$. Now, we can directly apply this limit to our transformed expression.

Step 4: Evaluate the limit

Therefore, $\lim_{u \rightarrow 0} \frac{\sin(u)}{u} = 1$. This straightforward application of the special limit provides the final answer. The simplicity of this step underscores the effectiveness of the substitution technique in transforming the original problem into a readily solvable form.

Thus, $\lim _{x \rightarrow \infty} x \sin \left(\frac{1}{x}\right) = 1$. This result demonstrates how a well-chosen substitution can simplify a seemingly complex limit and allow us to apply known limit properties.

Conclusion

In this comprehensive guide, we have successfully evaluated two distinct limits, each requiring a unique approach. For $\lim _{x \rightarrow 1}\left(\frac{\sin (x-1)}{x^2+x-2}\right)$, we employed factoring, algebraic manipulation, and the special trigonometric limit to arrive at the answer of $\frac{1}{3}$. For $\lim _{x \rightarrow \infty} x \sin \left(\frac{1}{x}\right)$, we utilized a clever substitution to transform the limit into a form where the special trigonometric limit could be directly applied, yielding a result of 1. These examples underscore the importance of mastering various techniques for limit evaluation, including algebraic manipulation, trigonometric identities, and strategic substitutions. By understanding these principles and practicing their application, you can confidently navigate the world of limits and unlock the doors to more advanced calculus concepts. Mastering limits is not just about finding answers; it's about developing a deep understanding of how functions behave and interact, which is essential for success in calculus and beyond. Remember, the journey to mathematical mastery is paved with practice and perseverance. So, continue to explore, experiment, and challenge yourself with new and exciting limit problems!

By understanding these principles and practicing their application, you can confidently navigate the world of limits and unlock the doors to more advanced calculus concepts. Mastering limits is not just about finding answers; it's about developing a deep understanding of how functions behave and interact, which is essential for success in calculus and beyond. Remember, the journey to mathematical mastery is paved with practice and perseverance. So, continue to explore, experiment, and challenge yourself with new and exciting limit problems!