Piecewise function continuity is a crucial concept in calculus, and understanding how to ensure continuity in these functions is essential. This post delves into a specific example, focusing on finding the value of ‘a’ that guarantees continuity at a specific point for a piecewise function. We’ll actively explore the necessary steps to determine this critical value.

To ensure piecewise function continuity, we need to equate the limit of the function as ‘x’ approaches the point in question to the function’s value at that point. This involves evaluating the limit of the function’s expression for ‘x’ not equal to the point, and then setting it equal to the function’s value at the point. This process allows us to determine the missing piece, ‘a’, in our piecewise function, guaranteeing continuity. Ultimately, we’ll use Taylor series and L’Hôpital’s Rule (or similar techniques) to evaluate the limit and solve for ‘a’.

“Success is not final, failure is not fatal: it is the courage to continue that counts” – Winston Churchill

Piecewise Function Continuity

This post explores a crucial calculus concept: ensuring continuity of a piecewise function. Let’s examine the problem and find the value of ##a##.

Problem Statement

Given the piecewise function:

### f(x) = \begin{cases} \frac{e^x – 1}{x} & \text{if } x \neq 0, \\ a & \text{if } x = 0. \end{cases} ###

Find the value of ##a## that makes ##f(x)## continuous at ##x = 0##.

Solution

Understanding the Problem

For a function to be continuous at a point, the limit of the function as ##x## approaches that point must equal the function’s value at that point. In this case, we need to find the limit of ##f(x)## as ##x## approaches 0.

Solving the Problem

Step 1: Evaluating the Limit

We need to find ##\lim_{x \to 0} f(x)## for ##x \neq 0##:

### \lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{e^x – 1}{x} ###

This limit has the indeterminate form ##\frac{0}{0}##, so we can apply L’Hôpital’s Rule or use the Taylor series expansion for ##e^x##.

Step 2: Using Taylor Series

The Taylor series expansion for ##e^x## around ##x = 0## is:

### e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + … ###

Substituting this into the limit expression:

### \lim_{x \to 0} \frac{e^x – 1}{x} = \lim_{x \to 0} \frac{(1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + …) – 1}{x} = \lim_{x \to 0} \frac{x + \frac{x^2}{2!} + \frac{x^3}{3!} + …}{x} ###

Simplifying the expression:

### = \lim_{x \to 0} (1 + \frac{x}{2!} + \frac{x^2}{3!} + …) ###

As ##x## approaches 0, all terms with ##x## vanish, leaving:

### = 1 ###

Step 3: Determining ‘a’

For continuity, the limit must equal the function value at ##x = 0##:

### \lim_{x \to 0} f(x) = f(0) ###

Therefore, ##1 = a##.

Final Solution

The value of ##a## that ensures continuity is:

##a = 1##

This example highlights the importance of understanding limits and Taylor series expansions in analyzing the continuity of piecewise functions. This is a fundamental concept in Calculus.

Concept	Equation/Definition	Explanation
Piecewise Function	### f(x) = \begin{cases} \frac{e^x – 1}{x} & \text{if } x \neq 0, \\ a & \text{if } x = 0. \end{cases} ###	The function is defined differently for different values of x. Continuity at a specific point (x=0) is crucial for piecewise functions.
Continuity Condition	### lim_{x → 0} f(x) = f(0) ###	For a function to be continuous at a point, the limit of the function as x approaches that point must equal the function’s value at that point.
Limit Evaluation (x≠0)	### lim_{x → 0} (ex – 1) / x ###	This limit has the indeterminate form 0/0, requiring further analysis (L’Hôpital’s Rule or Taylor series).
Taylor Series for ex	### ex = 1 + x + x2/2! + x3/3! + … ###	Used to evaluate the limit by substituting the Taylor expansion of ex.
Limit Evaluation (using Taylor Series)	### lim_{x → 0} (1 + x + x2/2! + … – 1) / x = 1 ###	Simplifying the limit expression using the Taylor series reveals the limit is 1.
Value of ‘a’ for Continuity	a = 1	For continuity at x=0, the limit must equal the function value at x=0, thus a=1.
Piecewise Function Continuity	–	This example demonstrates how to ensure continuity of a piecewise function by finding the appropriate value for a parameter. The continuity of a piecewise function is essential in many areas of mathematics and physics.

This post meticulously demonstrates how to determine the value of ‘a’ that ensures continuity for a piecewise function. We’ve tackled a specific example, providing a clear step-by-step approach. Understanding piecewise function continuity is crucial for advanced calculus and its applications.

By equating the limit of the function as ‘x’ approaches the point of interest to the function’s value at that point, we can identify the missing piece, ‘a’. This involves evaluating the limit of the function’s expression for ‘x’ not equal to the point and then setting it equal to the function’s value at the point. This process guarantees continuity, a fundamental concept in calculus.

• Understanding the Problem: Piecewise function continuity hinges on the limit’s equality to the function’s value at the point.
• Solving the Problem: We’ve used Taylor series expansions and L’Hôpital’s Rule (or similar techniques) to find the limit.
• Determining ‘a’: The crucial step involves setting the limit equal to the function’s value at the point to solve for ‘a’.
• Practical Application: This concept is vital for understanding and working with more complex functions in calculus and related fields.
• Practical Application: This concept is vital for understanding and working with more complex functions in calculus and related fields.

This example underscores the importance of piecewise function continuity in calculus. The ability to identify and address discontinuities in functions is essential for various mathematical and scientific applications. By understanding these principles, students can tackle more intricate problems involving piecewise functions and their continuity properties.