Exploring the Limit of ((x^{2/3} - 1)/(x^{3/5} - 1)) as x Approaches 1

The problem of finding the limit of (frac{x^{2/3} - 1}{x^{3/5} - 1}) as (x) approaches 1 involves exploring indeterminate forms and the application of L'H?pital's Rule. This article will guide you through the process, including necessary calculus steps and the importance of proper notation.

Introduction to Calculus and Limits

In calculus, understanding limits is fundamental. Limits describe the behavior of a function as its input values approach a certain value. In this case, we are dealing with an indeterminate form, which can be resolved using L'H?pital's Rule.

Solution to the Limit Problem

Let's start by examining the limit in question:

[ lim_{x to 1} frac{x^{2/3} - 1}{x^{3/5} - 1} ]

Step 1: Check the form of the limit

As (x to 1):[ x^{2/3} - 1 to 0 ][ x^{3/5} - 1 to 0 ]This gives us the indeterminate form (frac{0}{0}).

Step 2: Apply L'H?pital's Rule

According to L'H?pital's Rule, if the limit of the ratio of the derivatives exists or is an indeterminate form, we can differentiate the numerator and the denominator separately:

[ frac{d}{dx} left(x^{2/3} - 1right) frac{2}{3} x^{-1/3} ][ frac{d}{dx} left(x^{3/5} - 1right) frac{3}{5} x^{-2/5} ]

Step 3: Rewrite the limit using the derivatives

Applying L'H?pital's Rule, we rewrite the limit as:

[ lim_{x to 1} frac{frac{2}{3} x^{-1/3}}{frac{3}{5} x^{-2/5}} ]

Step 4: Simplify the expression

This simplifies to:

[ lim_{x to 1} frac{frac{2}{3}}{frac{3}{5}} cdot frac{x^{-1/3}}{x^{-2/5}} lim_{x to 1} frac{2}{3} cdot frac{5}{3} cdot x^{2/5 - 1/3} ]

Combining the exponents:

[ 2/5 - 1/3 frac{6 - 5}{15} frac{1}{15} ]

So the limit becomes:

[ lim_{x to 1} frac{10}{9} x^{1/15} ]

Step 5: Evaluate the limit

Substituting (x 1), we get:

[ frac{10}{9} cdot 1^{1/15} frac{10}{9} ]

Therefore, the limit is:

[ boxed{frac{10}{9}} ]

Additional Considerations and Alternative Approaches

It is important to note that there are alternative ways to approach such problems. In some cases, variable substitution can also be helpful. For instance, let (x u^{15}). As (x to 1), then (u to 1). Substituting (x u^{15}) into the original limit, we get:

[ lim_{x to 1} frac{u^{10} - 1}{u^9 - 1} ]

Factoring can further simplify this expression, leading to:

[ lim_{u to 1} frac{(u-1)(u^9 u^8 dots 1)}{(u-1)(u^8 u^7 dots 1)} ]

After canceling out the common factors, we are left with:

[ lim_{u to 1} frac{u^9 u^8 dots 1}{u^8 u^7 dots 1} ]

Evaluating the limit as (u to 1) gives:

[ frac{1 1 dots 1}{1 1 dots 1} frac{10}{9} ]

Therefore, using both methods, we confirm that the limit is indeed (boxed{frac{10}{9}}).

Conclusion

This limit problem showcases the importance of recognizing indeterminate forms and applying L'H?pital's Rule appropriately. The alternative approach through variable substitution further reinforces our understanding of the limit's behavior near a specific point. Understanding these techniques is crucial for solving more complex problems in calculus.