Continuity Analysis Of F(x) At X=0 And Discontinuity Type

Introduction

In this article, we delve into the continuity of the function f(x) defined piecewise as follows:

f(x) = \begin{cases} \frac{e^{1/x}}{1 + e^{1/x}}, & \text{if } x \neq 0 \\ 0, & \text{if } x = 0 \end{cases}

Our primary focus is to examine the continuity of this function at the point x = 0. This involves a rigorous analysis of the function's behavior as x approaches 0 from both the left and the right, and comparing these limits with the function's value at x = 0. Furthermore, if the function is found to be discontinuous, we will identify the specific type of discontinuity present. Understanding continuity is crucial in calculus as it forms the bedrock for many advanced concepts, including differentiability and integrability. A function is said to be continuous at a point if its limit exists at that point, the function is defined at that point, and the limit's value matches the function's value. If any of these conditions are not met, the function is discontinuous at that point. Discontinuities can be classified into several types, such as removable, jump, and essential discontinuities, each with distinct characteristics and implications for the function's behavior. This exploration will not only determine the continuity of the given function at x = 0 but also provide a deeper insight into the nature of its discontinuity, if any.

Understanding Continuity and Discontinuity

Before we dive into the specifics of the given function, it's important to understand the fundamental concepts of continuity and discontinuity. A function f(x) is said to be continuous at a point x = a if it satisfies three conditions: 1) f(a) is defined, 2) the limit of f(x) as x approaches a exists, and 3) the limit of f(x) as x approaches a is equal to f(a). Mathematically, this can be expressed as:

\lim_{x \to a} f(x) = f(a)

If any of these conditions are not met, the function is said to be discontinuous at x = a. There are several types of discontinuities, each with distinct characteristics. A removable discontinuity occurs when the limit of f(x) as x approaches a exists, but it is not equal to f(a), or f(a) is not defined. This type of discontinuity can be 'removed' by redefining the function at that point. A jump discontinuity occurs when the left-hand limit and the right-hand limit at x = a both exist but are not equal. This results in a 'jump' in the function's graph. An essential discontinuity (also known as an infinite discontinuity) occurs when at least one of the one-sided limits at x = a does not exist or is infinite. This type of discontinuity cannot be removed or 'fixed'. Understanding these types of discontinuities is essential for analyzing the behavior of functions and their applications in various fields. In the context of our function, we will be examining whether it meets the criteria for continuity at x = 0, and if not, we will classify the type of discontinuity present.

Evaluating the Limit as x Approaches 0 from the Left

To determine the continuity of the function f(x) at x = 0, we first need to evaluate the left-hand limit. This involves analyzing the behavior of f(x) as x approaches 0 from values less than 0. In other words, we are looking at the limit:

\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{e^{1/x}}{1 + e^{1/x}}

As x approaches 0 from the left, 1/x approaches negative infinity. Therefore, e^(1/x) approaches 0. This is a crucial observation because it allows us to simplify the expression. When e^(1/x) approaches 0, the numerator of the fraction approaches 0, and the denominator approaches 1 (since 1 + 0 = 1). Thus, the entire fraction approaches 0. Mathematically, we can express this as:

\lim_{x \to 0^-} \frac{e^{1/x}}{1 + e^{1/x}} = \frac{0}{1 + 0} = 0

This result indicates that the left-hand limit of f(x) as x approaches 0 exists and is equal to 0. However, this is only one part of the continuity analysis. We must also consider the right-hand limit and the function's value at x = 0 to make a conclusive determination about continuity. The left-hand limit provides a crucial piece of the puzzle, but it is not sufficient on its own to establish continuity. We need to compare it with the right-hand limit and the function's value at the point in question.

Evaluating the Limit as x Approaches 0 from the Right

Next, we need to evaluate the right-hand limit of the function f(x) as x approaches 0. This involves examining the behavior of f(x) as x approaches 0 from values greater than 0. The limit we are considering is:

\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \frac{e^{1/x}}{1 + e^{1/x}}

As x approaches 0 from the right, 1/x approaches positive infinity. Consequently, e^(1/x) approaches infinity. This situation requires a different approach than the left-hand limit. To handle the indeterminate form, we can divide both the numerator and the denominator by e^(1/x). This yields:

\lim_{x \to 0^+} \frac{e^{1/x}}{1 + e^{1/x}} = \lim_{x \to 0^+} \frac{1}{e^{-1/x} + 1}

Now, as x approaches 0 from the right, -1/x approaches negative infinity, and e^(-1/x) approaches 0. Therefore, the expression simplifies to:

\lim_{x \to 0^+} \frac{1}{e^{-1/x} + 1} = \frac{1}{0 + 1} = 1

This result shows that the right-hand limit of f(x) as x approaches 0 exists and is equal to 1. Comparing this with the left-hand limit, which we found to be 0, we observe that the two one-sided limits are not equal. This discrepancy is a key indicator of a discontinuity at x = 0. The unequal limits imply that the function 'jumps' at this point, which has significant implications for the continuity of f(x).

Determining the Function's Value at x = 0

To complete our continuity analysis, we need to determine the value of the function f(x) at x = 0. According to the piecewise definition of the function, we have:

f(0) = 0

This is directly given in the definition of f(x). Now that we have the left-hand limit, the right-hand limit, and the function's value at x = 0, we can compare these values to assess the continuity of f(x) at this point. We have established that:

• The left-hand limit as x approaches 0 is 0.
• The right-hand limit as x approaches 0 is 1.
• The function's value at x = 0 is 0.

By comparing these values, we can draw a definitive conclusion about the continuity of f(x) at x = 0. The next step is to synthesize these findings and identify the type of discontinuity present, if any.

Conclusion: Continuity Analysis and Type of Discontinuity

Having evaluated the left-hand limit, the right-hand limit, and the function's value at x = 0, we can now draw a conclusion about the continuity of f(x) at this point. We found that:

• The left-hand limit, ${\lim_{x \to 0^-} f(x) = 0}$.
• The right-hand limit, ${\lim_{x \to 0^+} f(x) = 1}$.
• The function's value at x = 0, f(0) = 0.

Since the left-hand limit and the right-hand limit are not equal (0 ≠ 1), the limit of f(x) as x approaches 0 does not exist. This immediately implies that the function is discontinuous at x = 0. Furthermore, because the left-hand and right-hand limits both exist but are different, we can classify this discontinuity as a jump discontinuity. A jump discontinuity is characterized by the function 'jumping' from one value to another at a particular point, which is precisely what we observe in this case. The function approaches 0 from the left and 1 from the right, creating a distinct jump at x = 0. In summary, the function f(x) is discontinuous at x = 0, and the type of discontinuity is a jump discontinuity. This analysis provides a complete understanding of the function's behavior around the point x = 0 and highlights the importance of considering both one-sided limits when assessing continuity.