Find The Limit Of The Function (√(x+1) - √2) / (x-1) As X Approaches 1.

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In the realm of calculus, evaluating limits is a fundamental concept. Limits allow us to analyze the behavior of functions as they approach specific values, including points where the function might be undefined. In this article, we will delve into the process of finding the limit of a radical function, specifically addressing the following problem:

lim⁡x→1x+13−23x−1\lim_{x\to 1} \frac{\sqrt[3]{x+1} - \sqrt[3]{2}}{x-1}

This limit presents an interesting challenge because direct substitution of x=1x = 1 results in the indeterminate form 00\frac{0}{0}. This signals that we need to employ algebraic manipulation techniques to simplify the expression and reveal the true limit. To effectively tackle this problem, we will explore the concepts of limits, indeterminate forms, and various algebraic techniques, including rationalization, that are commonly used to evaluate limits of complex functions. This step-by-step guide will not only provide a solution to the given limit but also equip you with the necessary tools to solve similar problems in calculus. Understanding these limit evaluation methods is crucial for grasping the core ideas behind continuity, derivatives, and integrals, which are cornerstones of calculus and its applications in diverse fields such as physics, engineering, and economics. By the end of this discussion, you'll have a solid understanding of how to approach complex limit problems involving radicals and indeterminate forms.

Understanding the Limit Problem

Before diving into the solution, let's break down the problem and understand the underlying concepts. The problem asks us to find the limit of the function f(x)=x+13−23x−1f(x) = \frac{\sqrt[3]{x+1} - \sqrt[3]{2}}{x-1} as xx approaches 1. This means we want to determine the value that f(x)f(x) gets arbitrarily close to as xx gets closer and closer to 1, without actually reaching 1. The concept of a limit is central to calculus and allows us to analyze the behavior of functions near specific points, especially where the function may not be defined. Direct substitution is often the first approach to evaluating limits. However, in this case, substituting x=1x = 1 directly into the function results in:

1+13−231−1=23−230=00\frac{\sqrt[3]{1+1} - \sqrt[3]{2}}{1-1} = \frac{\sqrt[3]{2} - \sqrt[3]{2}}{0} = \frac{0}{0}

This is known as an indeterminate form. Indeterminate forms, such as 00\frac{0}{0}, ∞∞\frac{\infty}{\infty}, 0⋅∞0 \cdot \infty, and ∞−∞\infty - \infty, do not have a defined value and indicate that further manipulation is required to evaluate the limit. In such cases, algebraic techniques like factoring, rationalizing, or L'Hôpital's Rule are often employed to transform the expression into a form where the limit can be easily determined. The presence of the cube root in our function suggests that rationalization might be a suitable technique. This involves multiplying the numerator and denominator by a conjugate expression that will eliminate the cube roots. The goal is to simplify the expression and remove the indeterminate form, allowing us to find the limit using direct substitution or other methods. Understanding the concept of indeterminate forms and recognizing when they occur is a crucial skill in calculus. It guides us towards choosing the appropriate method for evaluating limits. Furthermore, the existence of a limit at a point is closely related to the concept of continuity. If a function has a limit at a point and the limit is equal to the function's value at that point, then the function is said to be continuous at that point.

Applying Algebraic Manipulation: Rationalization

Since direct substitution leads to the indeterminate form 00\frac{0}{0}, we need to manipulate the expression algebraically to evaluate the limit. The presence of cube roots in the numerator suggests using rationalization. The idea behind rationalization is to eliminate the radicals by multiplying the numerator and denominator by a suitable expression. In this case, we have a difference of cube roots in the numerator: x+13−23\sqrt[3]{x+1} - \sqrt[3]{2}. To rationalize this expression, we need to recall the identity:

a3−b3=(a−b)(a2+ab+b2)a^3 - b^3 = (a - b)(a^2 + ab + b^2)

We can identify a=x+13a = \sqrt[3]{x+1} and b=23b = \sqrt[3]{2}. Therefore, we will multiply both the numerator and the denominator by the conjugate expression a2+ab+b2a^2 + ab + b^2, which in our case is:

(x+13)2+x+13⋅23+(23)2(\sqrt[3]{x+1})^2 + \sqrt[3]{x+1}\cdot\sqrt[3]{2} + (\sqrt[3]{2})^2

This step is crucial for removing the cube roots from the numerator and simplifying the expression. By multiplying by the conjugate, we are essentially reversing the difference of cubes factorization. Let's perform the multiplication:

lim⁡x→1x+13−23x−1⋅(x+13)2+x+13⋅23+(23)2(x+13)2+x+13⋅23+(23)2\lim_{x\to 1} \frac{\sqrt[3]{x+1} - \sqrt[3]{2}}{x-1} \cdot \frac{(\sqrt[3]{x+1})^2 + \sqrt[3]{x+1}\cdot\sqrt[3]{2} + (\sqrt[3]{2})^2}{(\sqrt[3]{x+1})^2 + \sqrt[3]{x+1}\cdot\sqrt[3]{2} + (\sqrt[3]{2})^2}

The numerator now becomes:

(x+13)3−(23)3=(x+1)−2=x−1(\sqrt[3]{x+1})^3 - (\sqrt[3]{2})^3 = (x+1) - 2 = x - 1

The expression now looks like this:

lim⁡x→1x−1(x−1)[(x+13)2+x+13⋅23+(23)2]\lim_{x\to 1} \frac{x-1}{(x-1)[(\sqrt[3]{x+1})^2 + \sqrt[3]{x+1}\cdot\sqrt[3]{2} + (\sqrt[3]{2})^2]}

Notice that we now have a factor of (x−1)(x-1) in both the numerator and the denominator. This is the key to removing the indeterminate form. This algebraic manipulation is a common technique used in limit evaluation when dealing with radicals. It highlights the importance of recognizing algebraic patterns and applying appropriate identities to simplify complex expressions. The choice of the conjugate expression is crucial for successfully rationalizing the numerator and eliminating the radical terms.

Simplifying and Evaluating the Limit

After rationalizing the numerator, we have the expression:

lim⁡x→1x−1(x−1)[(x+13)2+x+13⋅23+(23)2]\lim_{x\to 1} \frac{x-1}{(x-1)[(\sqrt[3]{x+1})^2 + \sqrt[3]{x+1}\cdot\sqrt[3]{2} + (\sqrt[3]{2})^2]}

We can now cancel the (x−1)(x-1) term from the numerator and denominator, as long as x≠1x \neq 1. This is valid because we are taking the limit as xx approaches 1, not at x=1x = 1 itself. Canceling the (x−1)(x-1) term removes the indeterminate form and simplifies the expression significantly. This step is a direct application of the properties of limits and the concept of removable discontinuities. By canceling the common factor, we are essentially removing a hole in the function's graph at x=1x=1, allowing us to evaluate the limit by direct substitution. The simplified expression is:

lim⁡x→11(x+13)2+x+13⋅23+(23)2\lim_{x\to 1} \frac{1}{(\sqrt[3]{x+1})^2 + \sqrt[3]{x+1}\cdot\sqrt[3]{2} + (\sqrt[3]{2})^2}

Now, we can use direct substitution to evaluate the limit. Substitute x=1x = 1 into the simplified expression:

1(1+13)2+1+13⋅23+(23)2=1(23)2+23⋅23+(23)2\frac{1}{(\sqrt[3]{1+1})^2 + \sqrt[3]{1+1}\cdot\sqrt[3]{2} + (\sqrt[3]{2})^2} = \frac{1}{(\sqrt[3]{2})^2 + \sqrt[3]{2}\cdot\sqrt[3]{2} + (\sqrt[3]{2})^2}

Simplifying further, we get:

1(23)2+(23)2+(23)2=13(23)2\frac{1}{(\sqrt[3]{2})^2 + (\sqrt[3]{2})^2 + (\sqrt[3]{2})^2} = \frac{1}{3(\sqrt[3]{2})^2}

We can rewrite (23)2(\sqrt[3]{2})^2 as 2232^{\frac{2}{3}}. So, the final answer is:

13⋅223\frac{1}{3 \cdot 2^{\frac{2}{3}}}

This is the value of the limit. We have successfully found the limit by using algebraic manipulation to remove the indeterminate form and then applying direct substitution. The process of simplifying and evaluating the limit involves a combination of algebraic skills and an understanding of the properties of limits. By carefully applying these techniques, we can determine the behavior of functions near points of discontinuity and find their limiting values.

Conclusion: Mastering Limit Evaluations

In this comprehensive guide, we successfully evaluated the limit:

lim⁡x→1x+13−23x−1=13⋅223\lim_{x\to 1} \frac{\sqrt[3]{x+1} - \sqrt[3]{2}}{x-1} = \frac{1}{3 \cdot 2^{\frac{2}{3}}}

We started by recognizing that direct substitution resulted in the indeterminate form 00\frac{0}{0}. This prompted us to employ algebraic manipulation techniques, specifically rationalization, to simplify the expression. By multiplying the numerator and denominator by the conjugate expression, we eliminated the cube roots and removed the indeterminate form. This example demonstrates the power of algebraic manipulation in evaluating limits. Recognizing patterns, applying appropriate identities, and simplifying expressions are crucial skills in calculus. After canceling the common factor (x−1)(x-1), we were able to use direct substitution to find the limit. This problem highlights the importance of understanding indeterminate forms and choosing the right strategy to handle them. Rationalization is a common technique for dealing with limits involving radicals, but other techniques, such as factoring and L'Hôpital's Rule, can be used in different situations. Mastering these limit evaluation techniques is essential for success in calculus and related fields. The ability to evaluate limits is fundamental to understanding continuity, derivatives, and integrals, which are core concepts in calculus and have wide-ranging applications in science, engineering, and economics. Furthermore, this problem emphasizes the importance of a step-by-step approach to problem-solving in mathematics. By carefully breaking down the problem, identifying the challenges, and applying appropriate techniques, we can arrive at the correct solution. This systematic approach is valuable not only in calculus but also in other areas of mathematics and beyond. By practicing and applying these techniques, you can build a solid foundation in calculus and enhance your problem-solving skills.