Understanding the Limit Problem: lim_{y->x} (y^{2/3} - x^{2/3}) / (y - x)

In calculus, understanding limits is crucial for both theoretical and practical applications. In this article, we will explore the limit problem: limx -> a (xp - ap) / (x - a). This concept can be generalized to a specific case: limy->x (y2/3 - x2/3) / (y - x).

Introduction to the Problem

The given limit problem involves the difference quotient, which is a fundamental concept in calculus. The difference quotient can be intuitively understood as the slope of the secant line between two points on a function's graph: (y2/3 - x2/3) / (y - x). The limit as y approaches x simplifies this to the derivative of the function at that point.

Applying the Power Rule

The power rule of differentiation states that for any function of the form f(x) xp, the derivative is given by:

limx -> a (xp - ap) / (x - a)  [d/dx (xp)]x  a  p ap - 1

This rule holds for any real value of p except when p 1. We will apply this rule to the specific case given: y2/3 - x2/3.

Simplifying the Expression

Let's start by recognizing that the expression can be simplified using the difference of cubes formula. The difference of cubes formula is given by:

a3 - b3 (a - b)(a2 ab b2)

In our case, we can rewrite:

y2/3 - x2/3 (y2/3 - x2/3)(y-1/3 y-4/3x1/3 y-1/3x2/3) / (y - x)

Now, let's apply this to our limit expression:

limy->x (y2/3 - x2/3) / (y - x)

Substituting the difference of cubes formula, we get:

limy->x ((y2/3 - x2/3)(y-1/3 y-4/3x1/3 y-1/3x2/3) / (y - x))

Since the term (y2/3 - x2/3) is common in the numerator and the expression (y - x) in the denominator, we can simplify it as:

limy->x (y-1/3 y-4/3x1/3 y-1/3x2/3)

Evaluating this as y approaches x, we get:

y-1/3 y-4/3x1/3 y-1/3x2/3 3x-1/3

Thus, the limit simplifies to:

limy->x (y2/3 - x2/3) / (y - x) -2/3x-1/3

Conclusion

The limit problem presented here showcases the application of basic differentiation rules, specifically the power rule. Understanding these concepts not only helps in solving complex problems but also solidifies the foundational knowledge of calculus. The specific limit problem discussed here highlights the elegance of calculus and the power of the limit concept in simplifying and solving seemingly complex expressions.