📐 Level 1 - Topic 4: Continuity: When Functions Flow Smoothly 🚀

1) Introduction: What Does it Mean for a Function to be "Continuous"?

Imagine tracing the graph of a function with your pencil. If you can trace the entire graph without lifting your pencil from the paper, the function is said to be continuous. Intuitively, continuity means there are no breaks, jumps, holes, or vertical asymptotes in the graph of the function at a particular point or over an interval.

Continuity is a fundamental concept in calculus and has far-reaching implications. Continuous functions behave predictably and smoothly, which is essential for many theorems and applications in calculus, physics, engineering, and economics. For example, optimization problems, root finding, and many physical models rely heavily on the property of continuity.

Continuity and Limits: A Tight Connection

Continuity is rigorously defined using the concept of limits we explored in the previous topics. In essence, continuity at a point \( x = a \) means that the limit of the function as \( x \) approaches \( a \) exists and is equal to the function's value at \( a \). This connection is crucial for a precise understanding of continuity.

2) The Formal Definition of Continuity at a Point

For a function \( f(x) \) to be continuous at a point \( x = a \), three conditions must be met simultaneously. This is the rigorous definition that ensures no "breaks, jumps, or holes" exist at \( x = a \).

Definition: Continuity at a Point \( x = a \)

A function \( f(x) \) is said to be continuous at \( x = a \) if and only if all three of the following conditions are satisfied:

Condition 1: \( f(a) \) is defined.
This means that the function must have a value at \( x = a \). In other words, \( a \) must be in the domain of \( f \). We cannot talk about continuity at a point if the function is not even defined there.
Condition 2: \( \lim_{x \to a} f(x) \) exists.
The limit of \( f(x) \) as \( x \) approaches \( a \) must exist. This implies that both the left-hand limit \( \lim_{x \to a^-} f(x) \) and the right-hand limit \( \lim_{x \to a^+} f(x) \) must exist and be equal to the same finite number. If the limit does not exist (e.g., due to differing one-sided limits, infinite limit, or oscillations), the function cannot be continuous at \( x = a \).
Condition 3: \( \lim_{x \to a} f(x) = f(a) \).
The limit of \( f(x) \) as \( x \) approaches \( a \) must be equal to the function's value at \( x = a \). This is the crucial condition that links the behavior of the function *near* \( a \) to its value *at* \( a \), ensuring a smooth transition and no gap.

If any one of these three conditions fails, then \( f(x) \) is discontinuous at \( x = a \).

Understanding the Three Conditions

Think of these conditions as a checklist. To verify continuity at a point, you must check all three. If even one condition is not met, continuity fails. Condition 1 ensures the function is defined. Condition 2 ensures the function is "approaching a single value" as you get closer to \( a \). Condition 3 is the glue that connects the limit (the approached value) to the actual function value at \( a \), ensuring no "jump" or "hole".

3) Types of Discontinuities: Recognizing Breaks in the Graph

When a function is discontinuous at a point, it's important to classify the *type* of discontinuity. Understanding the types helps us analyze function behavior and sometimes "fix" discontinuities where possible (removable discontinuities).

3.1) Removable Discontinuity (Hole)

A removable discontinuity occurs at \( x = a \) when \( \lim_{x \to a} f(x) \) exists, but either \( f(a) \) is not defined (Condition 1 fails) or \( f(a) \) is defined but \( \lim_{x \to a} f(x) \neq f(a) \) (Condition 3 fails). In this case, there is a "hole" in the graph at \( x = a \). We can "remove" this discontinuity by redefining or defining \( f(a) \) to be equal to \( \lim_{x \to a} f(x) \).

Example 1: Removable Discontinuity

Consider \( r(x) = \frac{x^2 - 4}{x - 2} \). Is \( r(x) \) continuous at \( x = 2 \)?

Condition 1: \( r(2) \) is defined? No, \( r(2) = \frac{2^2 - 4}{2 - 2} = \frac{0}{0} \) is undefined. Condition 1 fails.
Since Condition 1 fails, we already know \( r(x) \) is discontinuous at \( x = 2 \). Let's check the limit anyway: \( \lim_{x \to 2} \frac{x^2 - 4}{x - 2} = \lim_{x \to 2} \frac{(x - 2)(x + 2)}{x - 2} = \lim_{x \to 2} (x + 2) = 4 \). The limit exists.
Condition 3 is not applicable since Condition 1 fails.

\( r(x) \) has a removable discontinuity at \( x = 2 \). We could make it continuous by defining \( r(2) = 4 \).

3.2) Jump Discontinuity

A jump discontinuity occurs at \( x = a \) when the left-hand limit \( \lim_{x \to a^-} f(x) \) and the right-hand limit \( \lim_{x \to a^+} f(x) \) both exist, but are not equal, i.e., \( \lim_{x \to a^-} f(x) \neq \lim_{x \to a^+} f(x) \). In this case, \( \lim_{x \to a} f(x) \) does not exist (Condition 2 fails), leading to a "jump" in the graph at \( x = a \).

Example 2: Jump Discontinuity

Consider the piecewise function \( j(x) = \begin{cases} x + 1 & \text{if } x < 1 \\ 4 - x & \text{if } x \geq 1 \end{cases} \). Is \( j(x) \) continuous at \( x = 1 \)?

Condition 1: \( j(1) \) is defined? Yes, \( j(1) = 4 - 1 = 3 \).
Condition 2: \( \lim_{x \to 1} j(x) \) exists?
- Left-hand limit: \( \lim_{x \to 1^-} j(x) = \lim_{x \to 1^-} (x + 1) = 1 + 1 = 2 \).
- Right-hand limit: \( \lim_{x \to 1^+} j(x) = \lim_{x \to 1^+} (4 - x) = 4 - 1 = 3 \).
Since \( \lim_{x \to 1^-} j(x) = 2 \neq 3 = \lim_{x \to 1^+} j(x) \), \( \lim_{x \to 1} j(x) \) does not exist. Condition 2 fails.
Condition 3 is not applicable since Condition 2 fails.

\( j(x) \) has a jump discontinuity at \( x = 1 \).

3.3) Infinite Discontinuity (Vertical Asymptote)

An infinite discontinuity occurs at \( x = a \) when at least one of the one-sided limits approaches infinity or negative infinity, i.e., \( \lim_{x \to a^-} f(x) = \pm \infty \) or \( \lim_{x \to a^+} f(x) = \pm \infty \). In this case, \( \lim_{x \to a} f(x) \) does not exist (Condition 2 fails), and the function has a vertical asymptote at \( x = a \).

Example 3: Infinite Discontinuity

Consider \( v(x) = \frac{1}{x - 3} \). Is \( v(x) \) continuous at \( x = 3 \)?

Condition 1: \( v(3) \) is defined? No, \( v(3) = \frac{1}{3 - 3} = \frac{1}{0} \) is undefined. Condition 1 fails.
Since Condition 1 fails, we know \( v(x) \) is discontinuous at \( x = 3 \). Let's examine the limits:
- Right-hand limit: \( \lim_{x \to 3^+} \frac{1}{x - 3} = \infty \) (as \( x \) approaches 3 from the right, \( x - 3 \) is a small positive number).
- Left-hand limit: \( \lim_{x \to 3^-} \frac{1}{x - 3} = -\infty \) (as \( x \) approaches 3 from the left, \( x - 3 \) is a small negative number).
\( \lim_{x \to 3} v(x) \) does not exist (Condition 2 fails), and we have infinite discontinuity.
Condition 3 is not applicable since Condition 1 and 2 fail.

\( v(x) \) has an infinite discontinuity at \( x = 3 \), with a vertical asymptote at \( x = 3 \).

3.4) Oscillatory Discontinuity (Less Common at this Level, but for Completeness)

An oscillatory discontinuity occurs when the function values oscillate rapidly as \( x \) approaches \( a \), preventing the limit \( \lim_{x \to a} f(x) \) from existing. A classic example is \( f(x) = \sin(1/x) \) as \( x \to 0 \). These are more complex discontinuities.

Example 4: Oscillatory Discontinuity

Consider \( o(x) = \sin\left(\frac{1}{x}\right) \). Is \( o(x) \) continuous at \( x = 0 \)?

Condition 1: \( o(0) \) is defined? No, \( o(0) = \sin(1/0) \) is undefined. Condition 1 fails.
Even if we *were* to define \( o(0) \), \( \lim_{x \to 0} \sin\left(\frac{1}{x}\right) \) does not exist due to rapid oscillations between -1 and 1 as \( x \to 0 \). Condition 2 fails.
Condition 3 is not applicable.

\( o(x) \) has an oscillatory discontinuity at \( x = 0 \).

4) Continuity on an Interval: Extending Pointwise Continuity

We can extend the idea of continuity at a point to continuity over an interval.

Definition: Continuity on an Interval

Continuity on an Open Interval \( (a, b) \): A function \( f(x) \) is continuous on the open interval \( (a, b) \) if it is continuous at every point \( c \) in \( (a, b) \).
Continuity on a Closed Interval \( [a, b] \): A function \( f(x) \) is continuous on the closed interval \( [a, b] \) if:
1. It is continuous on the open interval \( (a, b) \).
2. It is continuous from the right at \( a \): \( \lim_{x \to a^+} f(x) = f(a) \).
3. It is continuous from the left at \( b \): \( \lim_{x \to b^-} f(x) = f(b) \).
For closed intervals, we need to consider one-sided continuity at the endpoints to ensure the function is "smoothly connected" up to and including the endpoints.

5) Properties of Continuous Functions: Building More Complex Continuous Functions

Continuous functions have some very useful properties. If we know certain functions are continuous, we can use these properties to determine the continuity of more complex functions built from them.

Properties of Continuous Functions

Let \( f(x) \) and \( g(x) \) be continuous at \( x = a \), and let \( c \) be a constant. Then the following functions are also continuous at \( x = a \):

Sum: \( f(x) + g(x) \)
Difference: \( f(x) - g(x) \)
Constant Multiple: \( c \cdot f(x) \)
Product: \( f(x) \cdot g(x) \)
Quotient: \( \frac{f(x)}{g(x)} \) (provided \( g(a) \neq 0 \))
Power: \( [f(x)]^n \), where \( n \) is a positive integer
Root: \( \sqrt[n]{f(x)} \) (for valid roots, and if \( n \) is even, assume \( f(a) \geq 0 \))
Composition: If \( g \) is continuous at \( a \) and \( f \) is continuous at \( g(a) \), then \( (f \circ g)(x) = f(g(x)) \) is continuous at \( a \).

Basic Building Block Functions are Continuous

Many elementary functions are continuous on their domains:

Polynomials are continuous everywhere (for all real numbers \( x \)).
Rational functions are continuous everywhere except where the denominator is zero (i.e., except at vertical asymptotes and removable discontinuities).
Trigonometric functions (\( \sin(x), \cos(x) \)) are continuous everywhere. Tangent, secant, cosecant, cotangent are continuous on their respective domains.
Exponential functions \( (a^x, a>0) \) are continuous everywhere.
Logarithmic functions \( (\log_a(x), a>0, a\neq 1) \) are continuous on their domains \( (x > 0) \).
Root functions \( (\sqrt[n]{x}) \) are continuous on their domains.
Absolute value function \( (|x|) \) is continuous everywhere.

Using these and the properties of continuity, we can determine the continuity of a vast array of functions.

6) Examples: Determining Continuity and Identifying Discontinuities

Let's examine several examples to practice determining continuity and identifying types of discontinuities.

Example 5: Polynomial - Continuous Everywhere

Is \( p(x) = x^3 - 5x^2 + 2x - 3 \) continuous at \( x = 2 \)?

\( p(2) = (2)^3 - 5(2)^2 + 2(2) - 3 = 8 - 20 + 4 - 3 = -11 \). Defined.
\( \lim_{x \to 2} p(x) \). Since \( p(x) \) is a polynomial, we can use direct substitution: \( \lim_{x \to 2} (x^3 - 5x^2 + 2x - 3) = (2)^3 - 5(2)^2 + 2(2) - 3 = -11 \). Limit exists.
\( \lim_{x \to 2} p(x) = p(2) = -11 \). Condition 3 is satisfied.

Yes, \( p(x) \) is continuous at \( x = 2 \). In fact, polynomials are continuous everywhere.

Example 6: Rational Function - Discontinuity at Zero of Denominator

Is \( q(x) = \frac{x + 1}{x - 2} \) continuous at \( x = 2 \) and at \( x = 0 \)?

At \( x = 2 \):

\( q(2) = \frac{2 + 1}{2 - 2} = \frac{3}{0} \), undefined. Condition 1 fails. Discontinuous at \( x = 2 \). (Infinite discontinuity).

At \( x = 0 \):

\( q(0) = \frac{0 + 1}{0 - 2} = -\frac{1}{2} \). Defined.
\( \lim_{x \to 0} q(x) = \lim_{x \to 0} \frac{x + 1}{x - 2} = \frac{0 + 1}{0 - 2} = -\frac{1}{2} \) (using quotient law and direct substitution). Limit exists.
\( \lim_{x \to 0} q(x) = q(0) = -\frac{1}{2} \). Condition 3 is satisfied.

\( q(x) \) is continuous at \( x = 0 \) but discontinuous at \( x = 2 \).

Example 7: Piecewise Function - Checking Continuity at the Break Point

Is \( w(x) = \begin{cases} x^2 & \text{if } x \leq 1 \\ 2x - 1 & \text{if } x > 1 \end{cases} \) continuous at \( x = 1 \)?

\( w(1) = (1)^2 = 1 \) (using the rule for \( x \leq 1 \)). Defined.
\( \lim_{x \to 1} w(x) \):
- \( \lim_{x \to 1^-} w(x) = \lim_{x \to 1^-} x^2 = (1)^2 = 1 \).
- \( \lim_{x \to 1^+} w(x) = \lim_{x \to 1^+} (2x - 1) = 2(1) - 1 = 1 \).
Since \( \lim_{x \to 1^-} w(x) = \lim_{x \to 1^+} w(x) = 1 \), \( \lim_{x \to 1} w(x) = 1 \). Limit exists.
\( \lim_{x \to 1} w(x) = 1 \) and \( w(1) = 1 \). So, \( \lim_{x \to 1} w(x) = w(1) \). Condition 3 is satisfied.

Yes, \( w(x) \) is continuous at \( x = 1 \).

Now consider \( z(x) = \begin{cases} x^2 & \text{if } x \leq 1 \\ 3x - 1 & \text{if } x > 1 \end{cases} \) at \( x = 1 \).

\( z(1) = (1)^2 = 1 \). Defined.
\( \lim_{x \to 1} z(x) \):
- \( \lim_{x \to 1^-} z(x) = \lim_{x \to 1^-} x^2 = (1)^2 = 1 \).
- \( \lim_{x \to 1^+} z(x) = \lim_{x \to 1^+} (3x - 1) = 3(1) - 1 = 2 \).
Since \( \lim_{x \to 1^-} z(x) = 1 \neq 2 = \lim_{x \to 1^+} z(x) \), \( \lim_{x \to 1} z(x) \) does not exist. Condition 2 fails. Discontinuous at \( x = 1 \) (Jump Discontinuity).

Example 8: Trigonometric Function - Continuous

Is \( t(x) = \cos(x) \) continuous at \( x = \pi \)?

\( t(\pi) = \cos(\pi) = -1 \). Defined.
\( \lim_{x \to \pi} \cos(x) \). Cosine is continuous, so direct substitution works: \( \lim_{x \to \pi} \cos(x) = \cos(\pi) = -1 \). Limit exists.
\( \lim_{x \to \pi} \cos(x) = t(\pi) = -1 \). Condition 3 is satisfied.

Yes, \( t(x) = \cos(x) \) is continuous at \( x = \pi \). Trigonometric functions \( \sin(x) \) and \( \cos(x) \) are continuous everywhere.

7) Practice Questions - Mastering Continuity Check

Time to practice applying the definition of continuity and identifying discontinuities.

Fundamental Practice Questions

Instructions: For each function and point \( a \), determine if the function is continuous at \( x = a \). If discontinuous, identify the type of discontinuity.

Q1. \( f(x) = 2x^2 - 3x + 4 \) at \( a = 3 \).
Q2. \( g(x) = \frac{x + 5}{x - 5} \) at \( a = 5 \) and at \( a = 0 \).
Q3. \( h(x) = \begin{cases} x + 2 & \text{if } x \leq 0 \\ x^2 + 2 & \text{if } x > 0 \end{cases} \) at \( a = 0 \).
Q4. \( k(x) = \frac{x^2 - 9}{x + 3} \) at \( a = -3 \) and at \( a = 0 \).
Q5. \( m(x) = |x - 2| \) at \( a = 2 \).
Q6. \( n(x) = \frac{1}{x^2 + 1} \) at \( a = 1 \).
Q7. \( p(x) = \begin{cases} 4 - x & \text{if } x < 2 \\ x^2 - 2 & \text{if } x \geq 2 \end{cases} \) at \( a = 2 \).
Q8. \( q(x) = \tan(x) \) at \( a = \pi/2 \) and at \( a = 0 \).
Q9. \( r(x) = \sqrt{x - 4} \) at \( a = 4 \) and at \( a = 0 \). (Consider domain for roots).
Q10. \( s(x) = \begin{cases} \frac{\sin(x)}{x} & \text{if } x \neq 0 \\ 1 & \text{if } x = 0 \end{cases} \) at \( a = 0 \). (This is a preview of the "sinc" function, a very important function).

Challenging Practice Questions

Instructions: These problems require more critical thinking about continuity and its properties.

Q1. Find the value(s) of constant \( C \) such that the function \( F(x) = \begin{cases} C x^2 + 1 & \text{if } x \leq -1 \\ 2x - C & \text{if } x > -1 \end{cases} \) is continuous at \( x = -1 \).
Q2. For what values of \( x \) is the function \( G(x) = \frac{\sqrt{x + 1}}{x^2 - 4} \) continuous? (Consider domain and points where denominator is zero). Express your answer in interval notation.
Q3. Explain why the function \( H(x) = \frac{|x|}{x} \) is discontinuous at \( x = 0 \). What type of discontinuity is it?
Q4. Suppose \( f(x) \) and \( g(x) \) are both discontinuous at \( x = a \). Is it possible for \( f(x) + g(x) \) to be continuous at \( x = a \)? If yes, provide an example. Is it always possible?
Q5. Consider a real-world scenario that can be modeled by a function. Describe two different scenarios: one where the function is continuous, and another where it is discontinuous at a specific point. Explain what continuity (or discontinuity) means in the context of these scenarios. (Think of situations involving smooth processes vs. abrupt changes).

8) Summary & Cheat Sheet - Continuity at a Glance

Let's recap the essential ideas about continuity for quick review and reference.

8.1) The Three Conditions for Continuity at \( x = a \)

For \( f(x) \) to be continuous at \( x = a \), all must hold:

\( f(a) \) is defined.
\( \lim_{x \to a} f(x) \) exists. ( \( \lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x) \))
\( \lim_{x \to a} f(x) = f(a) \).

8.2) Types of Discontinuities

Removable Discontinuity (Hole): Limit exists, but \( f(a) \) is undefined or \( \lim_{x \to a} f(x) \neq f(a) \).
Jump Discontinuity: One-sided limits exist but are unequal: \( \lim_{x \to a^-} f(x) \neq \lim_{x \to a^+} f(x) \).
Infinite Discontinuity (Vertical Asymptote): At least one one-sided limit is \( \pm \infty \).
Oscillatory Discontinuity: Function oscillates too rapidly near \( x = a \) for limit to exist.

8.3) Properties of Continuous Functions

If \( f, g \) are continuous at \( a \), then so are: \( f+g, f-g, c \cdot f, f \cdot g, f/g \) (if \( g(a) \neq 0 \)), \( [f]^n, \sqrt[n]{f} \) (valid roots), and \( f(g(x)) \) (composition). Basic functions like polynomials, \( \sin(x), \cos(x), a^x, \log(x), \sqrt[n]{x}, |x| \) are continuous on their domains.