The Limit of $(x - sin{x})/x^3$ as $x$ Approaches 0: Techniques and Proofs

In this article, we will explore the limit of the expression $(x - sin{x})/x^3$ as $x$ approaches 0. This is a classic problem that highlights the use of L'H?pital's Rule and Taylor Series in calculus. We will present and compare several methods to find the limit, emphasizing rigor and clarity.

Introduction

The limit of the form $(x - sin{x})/x^3$ as $x$ approaches 0 is interesting because it involves trigonometric functions and polynomials. Not only does it test the understanding of limit theorems, but it also sheds light on the behavior of these functions near zero. Understanding this limit can be crucial in various areas of calculus and mathematical analysis.

Solving the Limit Using L'H?pital's Rule

To solve the limit, we can use L'H?pital's Rule, which is applicable when the limit of a quotient of two functions is of the form $0/0$ or $pminfty/pminfty$. In our case, as $x$ approaches 0, both the numerator and the denominator approach 0, making it a suitable candidate for L'H?pital's Rule.

Step-by-Step Solution Using L'H?pital's Rule

Given the limit:

begin{equation}label{eq1}lim_{x to 0} frac{x - sin{x}}{x^3} frac{1}{6}.end{equation} Step 1: Start by applying L'H?pital's Rule: begin{align*}lim_{x to 0} frac{x - sin{x}}{x^3} lim_{x to 0} frac{1 - cos{x}}{3x^2} lim_{x to 0} frac{sin{x}}{6x} lim_{x to 0} frac{cos{x}}{6} frac{1}{6}.end{align*} Step 2: Alternatively, we can differentiate the numerator and the denominator repeatedly until the indeterminate form is resolved.

Through repeated application of L'H?pital's Rule, the limit can be shown to exist and is equal to $1/6$.

Solving the Limit Using Taylor Series Expansion

Another method to solve this limit involves the use of Taylor series expansion. Taylor series expansions provide a powerful tool to approximate functions, especially at a point of interest like $x 0$.

Step-by-Step Solution Using Taylor Series

Given that:

begin{equation}sin{x} x - frac{x^3}{3!} frac{x^5}{5!} - cdots.end{equation}

We can substitute this into the expression:

begin{align*}frac{x - sin{x}}{x^3} frac{x - left(x - frac{x^3}{3!} frac{x^5}{5!} - cdotsright)}{x^3} frac{frac{x^3}{3!} - frac{x^5}{5!} cdots}{x^3} frac{1}{3!} - frac{x^2}{5!} cdots.end{align*}

As $x$ approaches 0, all the higher-order terms (i.e., $x^2, x^3, ldots$) go to 0, leaving us with:

begin{equation}frac{1}{3!} frac{1}{6}.end{equation}

Hence, the limit is indeed $1/6$.

Alternative Solutions and Methods

There are other methods to solve this limit, and here are a couple more:

Dr. Sittinger's Solution

Dr. Sittinger’s approach involves a series transformation which is beyond the scope of this explanation but can be explored in further detail in mathematical literature.

Using Series Expansion and Limit Properties

We can use the fact that:

begin{equation}sin{x} x - frac{x^3}{6} O(x^5) quad text{as } x to 0.end{equation}

Thus, the expression becomes:

begin{align*}frac{x - sin{x}}{x^3} frac{x - (x - frac{x^3}{6} O(x^5))}{x^3} frac{frac{x^3}{6} O(x^5)}{x^3} frac{1}{6} O(x^2).end{align*}

As $x$ approaches 0, the higher-order terms vanish, leaving us with:

begin{equation}lim_{x to 0} frac{x - sin{x}}{x^3} frac{1}{6}.end{equation}

Conclusion

In conclusion, the limit of $(x - sin{x})/x^3$ as $x$ approaches 0 is a valuable exercise in understanding limits and their evaluation techniques. Whether using L'H?pital's Rule, Taylor series expansions, or alternative methods, the result is consistently $1/6$. This problem not only strengthens one's calculus skills but also deepens the understanding of fundamental concepts in mathematics.

References:

L'H?pital's Rule Taylor Series Expansion Dr. Sittinger’s Solution