Why is the Function (frac{sin x}{x}) Continuous at (x 0) When It Is Not Defined at (x 0)?

The question of the continuity of the function (f(x) frac{sin x}{x}) at (x 0) is a common point of confusion for many mathematics enthusiasts and students. This article delves into the nuances of this concept, providing a detailed explanation and addressing various viewpoints.

Understanding the Definition of Continuity

Continuity at a point for a function (f(x)) means that (lim_{x to a} f(x) f(a)). In the case of (f(x) frac{sin x}{x}), the function is not defined at (x 0), making it impossible to directly apply this definition at that point. However, the existence of a limit can help us determine the behavior of the function near this point.

The Limit as (x) Approaches (0)

Mathematically, we have:

[lim_{x to 0} frac{sin x}{x} 1]

This limit can be shown using L'H?pital's rule or, more intuitively, by examining the behavior of the sine function relative to (x) as (x) approaches zero. The limit existing means that the function (frac{sin x}{x}) approaches a finite value as (x) gets arbitrarily close to zero, suggesting a removable discontinuity.

Removable Discontinuity and Prolongation

When a function has a limit at a point but is not defined there, the discontinuity is removable. This means that the function can be made continuous at that point by redefining it appropriately. Specifically, if we define:

[f(0) 1]

then the function (f(x) frac{sin x}{x}) becomes continuous at (x 0). This redefinition ensures that the limit and the function value are equal at (x 0), fulfilling the condition for continuity.

Power Series Expansion

To further solidify our understanding, consider the power series expansion of (sin x). The Taylor series expansion of (sin x) around (x 0) is:

[sin x x - frac{x^3}{3!} frac{x^5}{5!} - frac{x^7}{7!} cdots]

Dividing both sides by (x), we get:

[frac{sin x}{x} 1 - frac{x^2}{3!} frac{x^4}{5!} - frac{x^6}{7!} cdots]

As (x) approaches zero, all terms involving higher powers of (x) approach zero, leaving:

[frac{sin x}{x} 1]

This power series approach again confirms the existence of the limit and the continuity at (x 0) when (f(0) 1).

Conclusion and Summary

In conclusion, while the function (f(x) frac{sin x}{x}) is not defined at (x 0), the limit (lim_{x to 0} frac{sin x}{x} 1) indicates a removable discontinuity. By redefining (f(0) 1), we can make the function continuous at (x 0). This redefinition ensures that the function matches the limit behavior at the origin, thus resolving the discontinuity.

Keywords: Continuity, Removable Discontinuity, Limit, Power Series Expansion