Continuity

In the realm of Mathematics, continuity is a fundamental concept that underpins various fields, from calculus to real analysis and beyond. At its core, continuity captures the notion of  unbrokenness in mathematical functions, providing a framework to understand the behavior of these functions over their domains.

Imagine tracing the path of a function on a graph without ever lifting your pen. If you can do so seamlessly, without encountering any sudden jumps or breaks, then that function is considered continuous. This idea of uninterrupted flow is not just visually appealing but also holds profound significance in mathematical reasoning and problem-solving.

In this exploration of continuity, we will delve into its definition, properties, and applications, unraveling its significance in the grand tapestry of mathematics. From the basics of continuous functions to advanced topics in analysis, join us on a journey to discover the seamless beauty of mathematical continuity.

1.0Define Continuity of a Function

Continuity of a function is a fundamental concept in mathematics that describes the connectedness of the graph of the function. Formally, a function f(x) is considered to be continuous at a point x = c if three conditions are met:

1. f(c) is defined, meaning that the function has a value at x = c.

2. The function has a defined limit as x approaches the value c. This means that as x gets arbitrarily close to c when the values of f(x) approach a specific finite value.

3. The value of the function at x = c coincides with the limit of the function at x = c, 

i.e., ${\lim }_{x\to c}$ f(x)=f(c).

In simpler terms, a function is continuous at a point if you can draw its graph without lifting your pen from the paper at that point. If the function meets these conditions for every point in its domain, it is considered continuous over its entire domain.

2.0Properties of Continuous Function

Continuous functions possess several key properties that make them essential tools in mathematical analysis and modeling. Here are some of the fundamental properties of continuous functions:

1. Existence of Limits: If a function f(x) demonstrates continuity at x = c, it implies that the limit of function f(x) as x approaches c exists and equals f(c).
2. Operations Preserving Continuity: The sum, difference, product, and quotient of two continuous functions are also continuous, provided the denominator is nonzero wherever it is defined.
3. Composition Preserving Continuity:  When f(x) exhibits continuity at x = c, and g(x) exhibits continuity at x = f(c), the composite function g(f(x) maintains continuity at x = c.
4. Intermediate Value Theorem: If f(x) is continuous over the closed interval [a, b], then there is guaranteed to be at least one value c within the interval [a, b] such that f (c) = y, where y falls between f(a) and f(b).
5. Extreme Value Theorem: If the function f(x) is continuous function over the closed interval [a, b], then f(x) attains both a maximum and a minimum value on [a, b].
6. Uniform Continuity: A function f(x) is uniformly continuous on a set S if, for every ε > 0, there exists a δ > 0 such that for all x, y in S, |x – y| < δ implies that |f(x) – f(y)| < ε, Uniformly continuous functions have a global property of continuity, unlike pointwise continuous functions.
7. Bolzano - Weierstrass Theorem: If function f(x) is continuous over a closed interval [a, b], then f(x) is bounded on [a, b] and attains its bounds.

These properties illustrate the power and versatility of continuous functions in mathematical analysis, providing a solid foundation for solving equations, studying the behavior of functions, and understanding the structure of mathematical objects.

3.0Continuity of Greatest Integer Function

The greatest integer function, denoted as (x), is a piecewise function that assigns to each real number x the greatest integer less than or equal to x. In other words, it rounds down to the nearest integer.

The continuity of the greatest integer function can be understood by examining its behavior around integers and between integers:

1. Around Integers: The function is discontinuous at integer values but continuous between them. For example, [2] = 2, [2.5] = 2 , and [3] = 3, but [2.9999] = 2. At integer values, there is a jump in the function's values, indicating discontinuity.
2. Between Integers: Between two consecutive integers n and n+1, the greatest integer function takes on the constant value n. Since there are no jumps between these values, the function is continuous over each interval between integers.

Formally, the greatest integer function is continuous on intervals of the form [n, n+1) and (n, n+1), where n is an integer, but it is discontinuous at integer values.

Graphically, the greatest integer function is represented by a series of horizontal line segments, each segment corresponding to a constant value over an interval, with jumps occurring at integer values.

In summary, the greatest integer function is continuous between integers but discontinuous at integer values. Its discontinuities occur as step functions, resulting in a piecewise continuous function.

4.0Solved Example on Continuity

Certainly! Let's work through another example:

Example 1: Determine whether the function f(x)=$\frac{x^2-4}{x-2}$ is continuous at x = 2.

Solution: To determine the continuity of f(x) at x = 2, we'll follow the same steps as before:

1. f (2) is defined: When x = 2, we have $f(2)=\frac{2^2-4}{2-2}=\frac{4-4}{0}$ _. However, division by zero is undefined, so f (2) is undefined.

2. exists: We'll find the limit of f(x) as x approaches 2.

   We can simplify $f(x)=\frac{x^2-4}{x-2}$ using factorization:

   $f(x)=\frac{(x+2)(x-2)}{x-2}$

   Notice that x – 2 cancels out, leaving f(x) = x + 2 for x ≠ 2. 

   Therefore,

${\lim }_{x\to 2}f(x)={\lim }_{x\to 2}$(x+2)= 2 + 2 = 4.

3. f(2) equals the limit ${\lim }_{x\to 2}f(x)$ : Since f(2) is undefined, it cannot be compared to the limit ${\lim }_{x\to 2}f(x)$ .

Since the limit ${\lim }_{x\to 2}$ f(x) exists but f(2) is undefined, we cannot apply the definition of continuity. The function f(x) = $\frac{x^2-4}{x-2}$ is not continuous at x = 2 due to the discontinuity caused by the division by zero.

This example demonstrates how to determine the continuity of a function at a specific point and highlights the importance of checking both the existence of the function's value and the limit as it approaches the point of interest. Let me know if you need further clarification or more examples!

Example 2: Determine the values of k for which the function f(x)=$\frac{2x+k}{x-3}$ ​ is continuous at x = 3.

Determine the values of k for which the function f(x)=$\{\begin{array}{cc}\frac{2x+k}{x-3};& x\ne 3\\ 2;& x=3\end{array}$ is continuous at x = 3.

Solution: For the function f(x) to be continuous at x = 3, both f(3) and ${\lim }_{x\to 3}$ ​f(x) must exist and be equal.

1. f (3): Substitute x = 3 into f(x):

f(3)= $\frac{2(3)+k}{3-3}=\frac{6+k}{0}$

For f (3) to exist, the denominator cannot be zero, so k must satisfy 6 + k ≠ 0.

2. ​ ${\lim }_{x\to 3}$f(x): Find the limit as x approaches 3.

${\lim }_{x\to 3}f(x)={\lim }_{x\to 3}\frac{2x+k}{x-3}$

 This limit exists if the function can be simplified or if L'Hôpital's Rule can be applied.

Since the limit exists only when k satisfies 6 + k ≠ 0, the values of k for which the function f(x) is continuous at x = 3 are all real numbers except for k = −6.

Example 3: Find the value of k for which the function f(x)= $\frac{x^2+kx+6}{x+2}$ is continuous for all real x.

Solution: For the function f(x) to be continuous for all real x, there must be no points of discontinuity, which means the denominator x + 2 cannot be zero.

x + 2 = 0 

x = −2

Therefore, the function f(x) will be continuous for all real x if x ≠ −2, which implies that the value of k does not affect the continuity of the function.

Example 4: Determine the value of “a” for which the function f(x)= $\frac{x^2-a^2}{x-a}$ is continuous at x = a.

Solution: For the function f(x) to be continuous at x = a, both f(a) and ${\lim }_{x\to a}f(x)$ must exist and be equal.

1. f(a): Substitute x = a into f(x):

f(a)= $\frac{a^2-a^2}{a-a}=\frac{0}{0}$

2. Find the limit as x approaches a.  ${\lim }_{x\to a}f(x)$

${\lim }_{x\to a}f(x)={\lim }_{x\to a}\frac{x^2-a^2}{x-a}$

This limit exists only if the function can be simplified or if L'Hôpital's Rule can be applied.

By factoring x² − a² as (x − a) (x + a), we get: 

${\lim }_{x\to a}f(x)={\lim }_{x\to a}\frac{(x-a)(x+a)}{x-a}={\lim }_{x\to a}(x+a)=2a$

For the function to be continuous at x = a, f(a) =${\lim }_{x\to a}$ ​f(x), so: 

0 = 2a 

a = 0

Therefore, the value of a for which the function f(x) is continuous at x = a is zero.

Example 5: Determine the value(s) of p for which the function f(x)= $\frac{x^2-5x+p}{x-3}$ is continuous for all real x.

Solution: For the function f(x) to be continuous for all real x, there must be no points of discontinuity, which means the denominator x – 3 cannot be zero.

x – 3 = 0 

x = 3 

Therefore, the function f(x) will be continuous for all real x if x ≠ 3, which implies that the value of p does not affect the continuity of the function.

Example 6: Determine the values of a for which the function f(x)= $\frac{3x-2a}{x-2}$ is continuous at x = 2.

Solution: For the function f(x) to be continuous at x = 2, both f (2) and ${\lim }_{x\to 2}$ f(x) must exist and be equal.

1. f(2): Substitute x = 2 into f(x):

$f(2)=\frac{3(2)-2a}{2-2}=\frac{6-2a}{0}$

Since the denominator is zero, f (2) is undefined.

2. Find the limit as x approaches 2.

${\lim }_{x\to 2}f(x)={\lim }_{x\to 2}\frac{3x-2a}{x-2}$

This limit exists only if the function can be simplified or if L'Hôpital's Rule can be applied.

By factoring the numerator:

${\lim }_{x\to 2}f(x)={\lim }_{x\to 2}\frac{(3x-6)(2a-4)}{x-2}$

$={\lim }_{x\to 2}\frac{3(x-2)-2(a-2)}{x-2}$

$={\lim }_{x\to 2}\left(3-\frac{2(a-2)}{x-2}\right)$

$={\lim }_{x\to 2}\left(3-\frac{2(a-2)}{x-2}\right)$

As x → 2, the term $\frac{2(a-2)}{x-2}$  approaches 2(a – 2). So, the limit becomes,

${\lim }_{x\to 2}(3-2(a-2))=3-2(a-2)=3-2a+4=7-2a$

For the function to be continuous at x = 2, ${\lim }_{x\to 2}f(x)$ is:

$\frac{6-2a}{0}=7-2a$

Since division by zero is undefined, the equation $\frac{6-2a}{0}=7-2a$ is not valid, and we cannot determine the values of a for which the function f(x) is continuous at x = 2.