How To Determine Continuity Of A Function

Continuity in mathematics, especially in calculus, is an intuitive concept that describes functions without abrupt changes or discontinuities. Understanding how to determine the continuity of a function is fundamental for calculus and real analysis. This article comprehensively explores the concept of continuity, provides step-by-step methods to determine whether a function is continuous, and discusses various types of discontinuities with illustrative examples.

Understanding Continuity

A function is said to be continuous at a point if its graph has no breaks, jumps, or holes at that point. More formally, a function f(x) is continuous at a point x = a if it satisfies the following three conditions:

1. f(a) is defined: The function must have a defined value at x = a.
2. Limit as x approaches a exists: The limit of f(x) as x approaches a must exist. This means that the left-hand limit and the right-hand limit at x = a must be equal.
3. Limit equals the function value: The limit of f(x) as x approaches a must be equal to f(a).

If any of these conditions are not met, the function is discontinuous at x = a.

Prerequisites

Before diving into the methods, ensure you have a solid understanding of the following:

• Limits: The concept of a limit is central to understanding continuity.
• Functions: Familiarity with different types of functions (polynomial, rational, trigonometric, exponential, logarithmic) is necessary.
• Algebra: Basic algebraic manipulations are often required to simplify and evaluate functions.

Steps to Determine Continuity of a Function

Here are detailed steps to determine whether a function f(x) is continuous at a point x = a:

Step 1: Check if f(a) is Defined

The first condition for continuity is that the function f(x) must be defined at the point x = a. In other words, when you substitute x = a into the function, you should get a real number as the output.

• Example 1:

  Consider the function f(x) = (x^2 - 4) / (x - 2) at x = 2.

  If we directly substitute x = 2, we get f(2) = (2^2 - 4) / (2 - 2) = 0 / 0, which is undefined. Therefore, f(x) is not continuous at x = 2.

• Example 2:

  Consider the function f(x) = x^2 + 3 at x = 1.

  Substituting x = 1, we get f(1) = (1)^2 + 3 = 4, which is defined. Thus, this condition is met.

Step 2: Evaluate the Limit as x Approaches a

The second condition requires that the limit of f(x) as x approaches a must exist. This involves checking both the left-hand limit (as x approaches a from the left) and the right-hand limit (as x approaches a from the right).

• The left-hand limit is denoted as lim x→a- f(x).
• The right-hand limit is denoted as lim x→a+ f(x).

For the limit to exist, these two one-sided limits must be equal.

• Example 1 (Continued):

  For f(x) = (x^2 - 4) / (x - 2) at x = 2, we need to evaluate the left-hand limit and the right-hand limit.

  • Left-hand limit: lim x→2- (x^2 - 4) / (x - 2) = lim x→2- (x + 2) = 4
  • Right-hand limit: lim x→2+ (x^2 - 4) / (x - 2) = lim x→2+ (x + 2) = 4

  Since both limits are equal, the limit exists and is equal to 4.

• Example 3:

  Consider the piecewise function:

  f(x) =

  {

  x + 1, if x < 2

  2x - 1, if x ≥ 2

  }

  at x = 2.

  • Left-hand limit: lim x→2- (x + 1) = 2 + 1 = 3
  • Right-hand limit: lim x→2+ (2x - 1) = 2(2) - 1 = 3

  Since both limits are equal, the limit exists and is equal to 3.

Step 3: Check if the Limit Equals the Function Value

The final condition is that the limit of f(x) as x approaches a must be equal to the value of the function at x = a. In other words, lim x→a f(x) = f(a).

• Example 1 (Continued):

  We found that lim x→2 (x^2 - 4) / (x - 2) = 4, but f(2) is undefined. Therefore, this function is not continuous at x = 2.

• Example 2 (Continued):

  We found that f(1) = 4. Now we evaluate the limit as x approaches 1:

  lim x→1 (x^2 + 3) = (1)^2 + 3 = 4

  Since lim x→1 f(x) = f(1) = 4, the function f(x) = x^2 + 3 is continuous at x = 1.

• Example 3 (Continued):

  We found that lim x→2 f(x) = 3. Now we check the value of the function at x = 2:

  f(2) = 2(2) - 1 = 3

  Since lim x→2 f(x) = f(2) = 3, the piecewise function is continuous at x = 2.

Summary of Steps

To summarize, determining continuity at a point x = a involves the following steps:

1. Check f(a) is Defined: Verify that the function has a defined value at x = a.
2. Evaluate the Limit:
   • Find the left-hand limit lim x→a- f(x).
   • Find the right-hand limit lim x→a+ f(x).
   • Ensure that lim x→a- f(x) = lim x→a+ f(x).
3. Compare Limit and Function Value: Confirm that lim x→a f(x) = f(a).

If all three conditions are met, the function is continuous at x = a.

Types of Discontinuities

If a function is not continuous at a point, it has a discontinuity. There are several types of discontinuities:

1. Removable Discontinuity

A removable discontinuity occurs when the limit of the function exists at a point, but the function is either undefined at that point or the value of the function at that point does not match the limit. This type of discontinuity can be "removed" by redefining the function at that point to match the limit.

• Example:

  f(x) = (x^2 - 4) / (x - 2) at x = 2.

  • lim x→2 (x^2 - 4) / (x - 2) = 4
  • f(2) is undefined.

  This is a removable discontinuity because we can redefine f(2) = 4 to make the function continuous at x = 2.

2. Jump Discontinuity

A jump discontinuity occurs when the left-hand limit and the right-hand limit exist at a point, but they are not equal. The function "jumps" from one value to another at that point.

• Example:

  Consider the piecewise function:

  f(x) =

  {

  x + 1, if x < 1

  x^2, if x ≥ 1

  }

  at x = 1.

  • Left-hand limit: lim x→1- (x + 1) = 2
  • Right-hand limit: lim x→1+ (x^2) = 1

  Since the left-hand limit and the right-hand limit are not equal, there is a jump discontinuity at x = 1.

3. Infinite Discontinuity

An infinite discontinuity occurs when the function approaches infinity (or negative infinity) as x approaches a certain value. This often happens with rational functions where the denominator approaches zero.

• Example:

  f(x) = 1 / x at x = 0.

  As x approaches 0, f(x) approaches infinity. Therefore, there is an infinite discontinuity at x = 0.

4. Oscillating Discontinuity

An oscillating discontinuity occurs when the function oscillates infinitely often near a point, preventing the limit from existing.

• Example:

  f(x) = sin(1 / x) at x = 0.

  As x approaches 0, the function sin(1 / x) oscillates infinitely between -1 and 1, and the limit does not exist.

Continuity on an Interval

A function is said to be continuous on an interval if it is continuous at every point within that interval. When dealing with closed intervals, the function must also be continuous from the right at the left endpoint and continuous from the left at the right endpoint.

Continuity from the Right

A function f(x) is continuous from the right at x = a if:

1. f(a) is defined.
2. The right-hand limit lim x→a+ f(x) exists.
3. lim x→a+ f(x) = f(a).

Continuity from the Left

A function f(x) is continuous from the left at x = a if:

1. f(a) is defined.
2. The left-hand limit lim x→a- f(x) exists.
3. lim x→a- f(x) = f(a).

Example

Consider the function f(x) = √x on the interval [0, ∞).

• For any a > 0, f(x) is continuous at x = a because the limit as x approaches a is equal to √a, which is f(a).
• At x = 0, we need to check continuity from the right:
  • f(0) = √0 = 0 is defined.
  • lim x→0+ √x = 0
  • Since lim x→0+ f(x) = f(0) = 0, the function is continuous from the right at x = 0.

Thus, f(x) = √x is continuous on the interval [0, ∞).

Theorems on Continuity

Several theorems help to determine the continuity of functions based on their composition and algebraic operations.

1. Continuity of Polynomials and Rational Functions

• Polynomials: All polynomial functions are continuous everywhere.
• Rational Functions: Rational functions (ratios of polynomials) are continuous everywhere except where the denominator is zero.

2. Continuity of Trigonometric Functions

• The sine and cosine functions are continuous everywhere.
• The tangent, cotangent, secant, and cosecant functions are continuous everywhere in their domains (i.e., where they are defined).

3. Continuity of Exponential and Logarithmic Functions

• Exponential functions (f(x) = a^x, where a > 0) are continuous everywhere.
• Logarithmic functions (f(x) = loga(x), where a > 0 and a ≠ 1) are continuous for x > 0.

4. Composition of Continuous Functions

If f(x) is continuous at x = a and g(x) is continuous at f(a), then the composite function g(f(x)) is continuous at x = a.

5. Algebraic Operations on Continuous Functions

If f(x) and g(x) are continuous at x = a, then the following functions are also continuous at x = a:

• f(x) + g(x)
• f(x) - g(x)
• c ⋅ f(x) (where c is a constant)
• f(x) ⋅ g(x)
• f(x) / g(x), provided that g(a) ≠ 0

Advanced Examples

Example 4: Continuity of a Piecewise Function

Determine whether the following piecewise function is continuous at x = 1 and x = 3:

f(x) =

{

x^2, if x ≤ 1

4 - x, if 1 < x < 3

x - 2, if x ≥ 3

}

At x = 1:

1. f(1) is defined: f(1) = (1)^2 = 1.
2. Evaluate the limit:
   • Left-hand limit: lim x→1- x^2 = (1)^2 = 1
   • Right-hand limit: lim x→1+ (4 - x) = 4 - 1 = 3 Since the left-hand limit and the right-hand limit are not equal, the limit does not exist at x = 1.

Therefore, the function is discontinuous at x = 1. Specifically, it has a jump discontinuity.

At x = 3:

1. f(3) is defined: f(3) = 3 - 2 = 1.
2. Evaluate the limit:
   • Left-hand limit: lim x→3- (4 - x) = 4 - 3 = 1
   • Right-hand limit: lim x→3+ (x - 2) = 3 - 2 = 1 Since the left-hand limit and the right-hand limit are equal, the limit exists and is equal to 1.
3. Compare limit and function value:
   • lim x→3 f(x) = 1 and f(3) = 1.

Since the limit exists and is equal to the function value, the function is continuous at x = 3.

Example 5: Using Continuity Theorems

Determine the intervals on which the function f(x) = (x^2 - 1) / √(x + 2) is continuous.

1. x^2 - 1 is a polynomial, so it is continuous everywhere.
2. √(x + 2) is continuous for x + 2 ≥ 0, which means x ≥ -2.
3. The quotient (x^2 - 1) / √(x + 2) is continuous wherever both functions are continuous and the denominator is not zero.
   • √(x + 2) = 0 when x = -2.

Therefore, the function is continuous on the interval (-2, ∞).

Conclusion

Determining the continuity of a function is a critical skill in calculus and real analysis. By understanding the three conditions for continuity and being able to identify different types of discontinuities, one can analyze the behavior of functions and their properties more effectively. The step-by-step methods and illustrative examples provided in this article offer a comprehensive guide to mastering this fundamental concept. Whether you are a student, educator, or professional, a solid grasp of continuity is essential for advanced mathematical studies and applications.