Solving A Limit Without L'Hopital's Rule: The Case of lim (x→0) (2x - xcosx - sinx) / x3

Understanding and solving limits is a fundamental aspect of calculus. While there are various methods to solve limits, L'Hopital's Rule is often a go-to tool. However, there are instances where it's more insightful and crafty to use series expansions and algebraic manipulations. In this article, we will explore a specific limit without relying on L'Hopital's Rule, focusing on the case lim (x→0) (2x - xcosx - sinx) / x3.

Introduction

The limit of the expression (2x - xcos x - sin x) / x3 as x approaches zero presents a challenge that can be solved through the use of Maclaurin series expansions. This approach provides a deeper understanding of the behavior of functions near specific points.

Maclaurin Series Expansions

A Maclaurin series is a Taylor series expansion around zero. The Maclaurin expansions for sine and cosine functions around zero are given by:

cos x 1 - frac{x^2}{2!} - frac{x^4}{4!} - frac{x^6}{6!} - ldots

sin x x - frac{x^3}{3!} - frac{x^5}{5!} - frac{x^7}{7!} - ldots

These series expansions are crucial for our analysis. We will use the first few terms of these series to approximate the behavior of the limit as x approaches zero.

Applying the Maclaurin Series

Let's substitute the Maclaurin expansions into the original limit:

2x - xcos x - sin x

Using the expansions, we get:

For cos x, we use the first few terms:

cos x 1 - frac{x^2}{2}

For sin x, we use:

sin x x - frac{x^3}{6}

Substituting these into our limit:

frac{2x - xleft(1 - frac{x^2}{2}right) - left(x - frac{x^3}{6}right)}{x^3} frac{2x - x frac{x^3}{2} - x frac{x^3}{6}}{x^3}

Simplifying the numerator:

2x - x - x frac{x^3}{2} frac{x^3}{6} -x frac{3x^3 x^3}{6} -x frac{4x^3}{6} -x frac{2x^3}{3}

So the expression simplifies to:

frac{-x frac{2x^3}{3}}{x^3} frac{-x}{x^3} frac{2x^3}{3x^3} frac{-1}{x^2} frac{2}{3}

As x approaches 0, the term frac{-1}{x^2} becomes unbounded, while frac{2}{3} remains. Therefore, the limit simplifies further to:

lim (x→0) frac{2x - xcos x - sin x}{x^3} frac{1}{6}.

Conclusion

The limit as x approaches zero of the expression (2x - xcos x - sin x) / x3 is indeed frac{1}{6}. This method not only provides a step-by-step resolution but also demonstrates the utility of Maclaurin series in analyzing intricate limits.

Additional Notes

Understanding such limits helps in advanced calculus and real analysis. Whether you're a student preparing for advanced mathematics, a professional in engineering, or a researcher in physics, knowing how to solve such limits without relying on L'Hopital's Rule can be particularly rewarding. It enhances one's problem-solving skills and deepens the intuition about the behavior of functions near critical points.

Resources available for further exploration include textbooks on calculus, online tutorials, and specific problem sets focusing on series expansions and limit problems. Practicing similar problems and understanding the underlying concepts will greatly enhance your mathematical toolkit.