How To Show A Function Is Continuous

Continuity of a function is a fundamental concept in calculus and real analysis, underpinning many advanced mathematical ideas. In essence, a function is continuous if its graph can be drawn without lifting your pen from the paper. This intuitive understanding translates into rigorous mathematical definitions that allow us to prove whether a function is continuous at a point or over its entire domain.

Defining Continuity: The Building Blocks

To rigorously demonstrate the continuity of a function, we first need to understand the formal definitions. Continuity can be examined at a single point or across an interval.

Continuity at a Point

A function f(x) is continuous at a point c if it satisfies three conditions:

1. f(c) is defined: The function must be defined at the point c. This means that c is in the domain of f.

2. The limit of f(x) as x approaches c exists: This implies that both the left-hand limit and the right-hand limit exist and are equal. Mathematically, this is expressed as:

   • lim (x→c-) f(x) = lim (x→c+) f(x)

3. The limit of f(x) as x approaches c is equal to f(c): The value that the function approaches as x approaches c must be the same as the function's value at c. This is the crucial link that ties the limit and the function value together:

   • lim (x→c) f(x) = f(c)

If any of these three conditions are not met, the function is said to be discontinuous at c.

Continuity on an Interval

A function f(x) is continuous on an open interval (a, b) if it is continuous at every point in that interval. For a function to be continuous on a closed interval [a, b], it must be continuous on the open interval (a, b), and it must also be continuous from the right at a and continuous from the left at b. This means:

• lim (x→a+) f(x) = f(a) (Continuity from the right at a)
• lim (x→b-) f(x) = f(b) (Continuity from the left at b)

These additional conditions ensure that the function behaves well at the endpoints of the interval.

Methods for Showing a Function is Continuous

There are several methods to prove the continuity of a function, each suited to different types of functions and situations. Here are some common approaches:

1. Direct Application of the Definition (Epsilon-Delta Proofs)

This is the most fundamental method and directly relies on the formal definition of a limit. It involves showing that for any arbitrary positive number ε (epsilon), there exists a positive number δ (delta) such that if x is within δ of c, then f(x) is within ε of f(c).

The Epsilon-Delta Definition of a Limit:

For every ε > 0, there exists a δ > 0 such that if 0 < |x - c| < δ, then |f(x) - f(c)| < ε.

Steps for an Epsilon-Delta Proof:

1. State the Goal: Clearly state what you need to show. For example, "We want to show that for every ε > 0, there exists a δ > 0 such that if 0 < |x - c| < δ, then |f(x) - f(c)| < ε."
2. "Scratch Work" (Finding Delta): This is the crucial step where you work backward from the inequality |f(x) - f(c)| < ε to find a suitable δ in terms of ε. This often involves algebraic manipulation and clever estimations. The goal is to find an expression for δ that depends on ε and c.
3. Formal Proof: Write out the proof formally. This involves:
   • Starting with an arbitrary ε > 0.
   • Stating the chosen value of δ (based on your scratch work).
   • Assuming 0 < |x - c| < δ.
   • Using this assumption and your chosen δ to show that |f(x) - f(c)| < ε.
4. Conclusion: Conclude that the limit exists and is equal to f(c), thus proving continuity at c.

Example: Proving f(x) = 2x + 1 is continuous at x = 2

1. Goal: Show that for every ε > 0, there exists a δ > 0 such that if 0 < |x - 2| < δ, then |(2x + 1) - (2(2) + 1)| < ε.

2. Scratch Work:

   • |(2x + 1) - (2(2) + 1)| = |2x + 1 - 5| = |2x - 4| = 2|x - 2|
   • We want 2|x - 2| < ε.
   • Dividing both sides by 2, we get |x - 2| < ε/2.
   • This suggests that we should choose δ = ε/2.

3. Formal Proof:

   • Let ε > 0 be given.
   • Choose δ = ε/2.
   • Assume 0 < |x - 2| < δ.
   • Then |(2x + 1) - 5| = |2x - 4| = 2|x - 2| < 2δ = 2(ε/2) = ε.
   • Therefore, |(2x + 1) - 5| < ε.

4. Conclusion: Since for every ε > 0, we found a δ > 0 such that if 0 < |x - 2| < δ, then |(2x + 1) - 5| < ε, we have shown that lim (x→2) (2x + 1) = 5 = f(2). Therefore, f(x) = 2x + 1 is continuous at x = 2.

Challenges and Considerations:

• Epsilon-delta proofs can be challenging, especially for more complex functions.
• Finding the appropriate δ often requires ingenuity and algebraic skill.
• These proofs are fundamental for understanding the definition of continuity but can be time-consuming.

2. Using Limit Laws and Properties of Continuous Functions

This method relies on established theorems about limits and continuity. It is often more efficient than epsilon-delta proofs for showing the continuity of combinations of simpler functions.

Key Theorems and Properties:

• Limits of Basic Functions: Limits of basic functions like polynomials, exponential functions, trigonometric functions (within their domains), and logarithmic functions (within their domains) are generally known and can be used directly. For example, lim (x→c) x = c and lim (x→c) k = k (where k is a constant).
• Limit Laws: These laws describe how limits interact with arithmetic operations:
  • lim (x→c) [f(x) + g(x)] = lim (x→c) f(x) + lim (x→c) g(x)
  • lim (x→c) [f(x) - g(x)] = lim (x→c) f(x) - lim (x→c) g(x)
  • lim (x→c) [k f(x)] = k lim (x→c) f(x)
  • lim (x→c) [f(x) g(x)] = lim (x→c) f(x) * lim (x→c) g(x)
  • lim (x→c) [f(x) / g(x)] = lim (x→c) f(x) / lim (x→c) g(x), provided lim (x→c) g(x) ≠ 0
• Continuity of Composite Functions: If g is continuous at c and f is continuous at g(c), then the composite function f(g(x)) is continuous at c.
• Continuity of Polynomials: All polynomial functions are continuous everywhere.
• Continuity of Rational Functions: Rational functions (ratios of polynomials) are continuous everywhere except where the denominator is zero.
• Intermediate Value Theorem: If f is continuous on the closed interval [a, b], and k is any number between f(a) and f(b), then there exists a number c in the interval [a, b] such that f(c) = k. While this theorem doesn't directly prove continuity, it relies on it and can be used to show the existence of solutions to equations.

Steps for Using Limit Laws and Properties:

1. Decompose the Function: Break down the given function into simpler functions whose continuity is known (e.g., polynomials, trigonometric functions, exponential functions).
2. Apply Limit Laws: Use the limit laws to evaluate the limit of the function as x approaches c.
3. Verify the Conditions: Check if f(c) is defined and if the limit as x approaches c is equal to f(c).
4. Invoke Continuity Properties: Use the properties of continuous functions (e.g., continuity of polynomials, rational functions, composite functions) to justify the continuity of the overall function.

Example: Proving f(x) = x^2 + 3x - 2 is continuous everywhere

1. Decompose the Function: f(x) = x^2 + 3x - 2 is a polynomial function.

2. Invoke Continuity Properties: We know that all polynomial functions are continuous everywhere.

3. Conclusion: Therefore, f(x) = x^2 + 3x - 2 is continuous everywhere. (This proof is very simple because we can directly apply a known property.)

Example: Proving f(x) = (x^2 - 1) / (x - 1) is continuous at x = 2

1. Decompose the Function: f(x) is a rational function.

2. Apply Limit Laws (with caution): We can't directly substitute x = 1 into the function because the denominator would be zero. However, we're interested in continuity at x = 2, where the denominator is not zero.

3. Simplify and Evaluate the Limit: For x ≠ 1, we can simplify the function: f(x) = (x^2 - 1) / (x - 1) = (x + 1)(x - 1) / (x - 1) = x + 1.

   • lim (x→2) f(x) = lim (x→2) (x + 1) = 2 + 1 = 3.

4. Verify the Conditions:

   • f(2) = (2^2 - 1) / (2 - 1) = 3 / 1 = 3.
   • lim (x→2) f(x) = 3 = f(2).

5. Invoke Continuity Properties: Since the simplified function x + 1 is a polynomial, it is continuous everywhere.

6. Conclusion: Therefore, f(x) = (x^2 - 1) / (x - 1) is continuous at x = 2. (Note that the function is not continuous at x = 1).

Advantages:

• Often more efficient than epsilon-delta proofs, especially for combinations of simpler functions.
• Relies on established theorems, making the proofs more concise and easier to understand.

Disadvantages:

• Requires familiarity with limit laws and properties of continuous functions.
• May not be applicable to all types of functions, especially those with piecewise definitions or singularities.

3. Using the Sequential Criterion for Continuity

The sequential criterion provides an alternative definition of continuity in terms of sequences.

The Sequential Criterion:

A function f(x) is continuous at a point c if and only if for every sequence {x_n} that converges to c, the sequence {f(x_n)} converges to f(c).

Steps for Using the Sequential Criterion:

1. Assume a Convergent Sequence: Let {x_n} be any sequence that converges to c. This means lim (n→∞) x_n = c.
2. Consider the Sequence of Function Values: Consider the sequence {f(x_n)}.
3. Show Convergence: Show that the sequence {f(x_n)} converges to f(c). This means you need to prove that lim (n→∞) f(x_n) = f(c). This may involve using limit laws, algebraic manipulation, or other techniques.
4. Conclusion: If you can show that for every sequence {x_n} converging to c, the sequence {f(x_n)} converges to f(c), then the function f(x) is continuous at c. If you can find one sequence {x_n} converging to c such that {f(x_n)} does not converge to f(c), then the function f(x) is discontinuous at c.

Example: Proving f(x) = x^2 is continuous at x = 3 using the Sequential Criterion

1. Assume a Convergent Sequence: Let {x_n} be any sequence that converges to 3. This means lim (n→∞) x_n = 3.

2. Consider the Sequence of Function Values: Consider the sequence {f(x_n)} = {x_n²}.

3. Show Convergence: We want to show that lim (n→∞) x_n² = f(3) = 3² = 9.

   • Since lim (n→∞) x_n = 3, we can use the limit law that states lim (n→∞) [f(x) g(x)] = lim (n→∞) f(x) * lim (n→∞) g(x).
   • Therefore, lim (n→∞) x_n² = lim (n→∞) (x_n * x_n) = lim (n→∞) x_n * lim (n→∞) x_n = 3 * 3 = 9.

4. Conclusion: Since for every sequence {x_n} converging to 3, the sequence {x_n²} converges to 9 = f(3), we have shown that f(x) = x² is continuous at x = 3.

Advantages:

• Provides a different perspective on continuity, linking it to the concept of sequences.
• Can be useful for proving the discontinuity of a function by finding a sequence that violates the criterion.

Disadvantages:

• May require a good understanding of sequences and their limits.
• Can be more abstract than the epsilon-delta definition.

4. Piecewise Functions: Special Considerations

Piecewise functions are defined by different formulas on different intervals. Showing the continuity of a piecewise function requires careful attention to the points where the function definition changes (the "breakpoints").

Steps for Showing Continuity of Piecewise Functions:

1. Check Continuity Within Each Interval: Verify that each piece of the function is continuous on its respective interval. This can often be done using limit laws and properties of continuous functions.
2. Check Continuity at the Breakpoints: For each breakpoint c, you need to check three things:
   • The left-hand limit exists: lim (x→c-) f(x) exists.
   • The right-hand limit exists: lim (x→c+) f(x) exists.
   • The limits are equal and equal to the function value at the breakpoint: lim (x→c-) f(x) = lim (x→c+) f(x) = f(c).
3. Consider Endpoint Continuity (for closed intervals): If the piecewise function is defined on a closed interval, you also need to check continuity from the right at the left endpoint and continuity from the left at the right endpoint.

Example:

Consider the piecewise function:

f(x) = { x², if x ≤ 1 { 2x - 1, if x > 1

Show that f(x) is continuous at x = 1.

1. Continuity Within Each Interval: x² is continuous for x ≤ 1, and 2x - 1 is continuous for x > 1 (since they are both polynomials).

2. Continuity at the Breakpoint (x = 1):

   • Left-hand limit: lim (x→1-) f(x) = lim (x→1-) x² = 1² = 1.
   • Right-hand limit: lim (x→1+) f(x) = lim (x→1+) (2x - 1) = 2(1) - 1 = 1.
   • Function value at x = 1: f(1) = 1² = 1.
   • Since lim (x→1-) f(x) = lim (x→1+) f(x) = f(1) = 1, the function is continuous at x = 1.

3. Conclusion: The piecewise function is continuous everywhere because each piece is continuous on its interval, and it is continuous at the breakpoint.

Key Point: The most common point of discontinuity in a piecewise function occurs at the breakpoints.

Examples of Continuous and Discontinuous Functions

• Continuous Functions:
  • Polynomials: f(x) = axⁿ + bx^n-1 + ... + c (for any constants a, b, c and non-negative integer n)
  • Exponential Functions: f(x) = a^x (for a > 0)
  • Sine and Cosine Functions: f(x) = sin(x) and f(x) = cos(x)
  • Constant Functions: f(x) = k (where k is a constant)
• Discontinuous Functions:
  • Rational Functions with Zero in the Denominator: f(x) = 1/x is discontinuous at x = 0.
  • Step Functions: The Heaviside step function (0 for x < 0, 1 for x ≥ 0) is discontinuous at x = 0.
  • Tangent Function: f(x) = tan(x) is discontinuous at x = π/2 + nπ (where n is an integer).
  • Piecewise Functions with Mismatched Limits: A piecewise function where the left-hand and right-hand limits at a breakpoint are not equal.

Why is Continuity Important?

Continuity is a crucial concept because it underlies many important theorems and applications in calculus and real analysis:

• Intermediate Value Theorem: As mentioned before, this theorem guarantees the existence of a value c such that f(c) = k for any k between f(a) and f(b), provided that f is continuous on [a, b].
• Extreme Value Theorem: If f is continuous on a closed interval [a, b], then f attains a maximum and a minimum value on that interval.
• Differentiability: While continuity does not guarantee differentiability, differentiability does imply continuity. If a function is differentiable at a point, it must also be continuous at that point. (The converse is not true; a function can be continuous but not differentiable, such as f(x) = |x| at x = 0).
• Integration: The definite integral of a continuous function exists.
• Applications in Physics and Engineering: Many physical phenomena are modeled by continuous functions. For example, the position of a moving object, the temperature of a room, and the voltage in an electrical circuit are often represented by continuous functions. Discontinuities in these models can represent sudden changes or idealizations.

Conclusion

Demonstrating the continuity of a function is a fundamental skill in calculus and analysis. Understanding the definition of continuity, mastering the epsilon-delta technique, utilizing limit laws and properties of continuous functions, applying the sequential criterion, and handling piecewise functions are all essential tools. While epsilon-delta proofs can be challenging, they provide a deep understanding of the definition of a limit. Limit laws and properties of continuous functions offer more efficient methods for proving continuity in many cases. The sequential criterion provides an alternative perspective and can be useful for proving discontinuity. Piecewise functions require careful attention at the breakpoints. Ultimately, mastering these techniques allows you to rigorously analyze the behavior of functions and apply them to a wide range of mathematical and real-world problems.