Chapter 1: Functions of Real Variables, Limits, Continuity, Graphs of a Function

Introduction: What is a Function?

Imagine a "Magic Box." You drop a number into the top, something happens inside, and a different number pops out the bottom. In mathematics, this "Magic Box" is called a Function.

A function is simply a rule that connects one number (the Input) to exactly one other number (the Output).

Key Rule: For every input, there can only be one output. If you put in a "2" and sometimes get a "4" and sometimes get a "10," it's not a function — it's a mess!

1. Functions of a Real Variable

When we say "Real Variable," we just mean that the numbers we are using are "Real Numbers" — basically any number you can find on a number line ($1$, $-5$, $0.5$, $\pi$, etc.).

We usually write functions like this:

$f(x) = y$

• $x$: The input (The Independent Variable).
• $f$: The rule (The "Magic Box").
• $y$: The output (The Dependent Variable).

2. Key Characteristics of Functions

A. Domain and Range

Domain: The set of all possible numbers you are allowed to plug in. (Example: You can't divide by zero, so $0$ would be excluded from the domain of $1/x$.)

Range: The set of all numbers that actually come out of the box.

3. Types of Functions

Type 1: Linear Functions (The Straight Line)

These are the simplest functions. They look like $f(x) = mx + c$.

Example 1: $f(x) = 2x + 3$. If $x = 1$, $f(1) = 2(1) + 3 = 5$.

Example 2: $f(x) = -x + 10$. If $x = 5$, $f(5) = -5 + 10 = 5$.

Type 2: Quadratic Functions (The U-Shape)

These have an $x^2$ in them. They create a curve called a parabola.

Example 3: $f(x) = x^2$. If $x = -3$, $f(-3) = (-3)^2 = 9$.

Example 4: $f(x) = x^2 - 4$. If $x = 2$, $f(2) = 0$.

Type 3: Polynomial Functions

These involve higher powers like $x^3$, $x^4$, etc.

Example 5: $f(x) = x^3 + 2x^2 - 1$.

Type 4: Rational Functions (The Fractions)

Functions where $x$ is in the denominator.

Example 6: $f(x) = \frac{1}{x}$. Note: $x$ cannot be $0$.

Type 5: Absolute Value Functions

These turn everything positive. $|-5| = 5$.

Example 7: $f(x) = |x|$. If $x = -10$, $f(x) = 10$.

4. More Advanced Types

Even and Odd Functions

Even: If you plug in a negative $x$, you get the same result as a positive $x$. $f(-x) = f(x)$. (Example: $x^2$).

Odd: If you plug in a negative $x$, you get the negative version of the result. $f(-x) = -f(x)$. (Example: $x^3$).

Composite Functions (The Double Box)

This is putting one function inside another. Written as $f(g(x))$.

Example 8: If $f(x) = x + 2$ and $g(x) = x^2$.

Then $f(g(3)) = f(3^2) = f(9) = 9 + 2 = 11$.

If we have two functions, $f$ and $g$, the composite function (written as $(f \circ g)(x)$ or $f(g(x))$) is defined as:

• Take the input $x$.
• Apply $g$ to it to get $g(x)$.
• Use the result $g(x)$ as the input for $f$.

Crucial Rule: The range of $g$ must be contained within the domain of $f$ for the composition to be defined.

5. The 5 Essential Problem Types (Step-by-Step)

Type 1: Basic Evaluation (Numerical Input)

Question: Given $f(x) = 2x + 3$ and $g(x) = x^2$, evaluate $(f \circ g)(4)$.

Step 1: Find the inner function value first: $g(4) = 4^2 = 16$.

Step 2: Substitute this result into the outer function $f$: $f(16) = 2(16) + 3$.

Final Answer: $2(16) + 3 = 32 + 3 = 35$.

Type 2: Algebraic Composition (Variable Input)

Question: Given $f(x) = \frac{1}{x+1}$ and $g(x) = 3x - 2$, find $(g \circ f)(x)$.

Step 1: Write the structure: $g(f(x))$.

Step 2: Substitute the entire expression for $f(x)$ into $g$: $g\left(\frac{1}{x+1}\right) = 3\left(\frac{1}{x+1}\right) - 2$.

Step 3: Simplify: $\frac{3}{x+1} - 2 = \frac{3 - 2(x+1)}{x+1} = \frac{3 - 2x - 2}{x+1}$.

Final Answer: $\frac{1 - 2x}{x+1}$.

Type 3: Finding the Domain of a Composite Function

Question: Find the domain of $(f \circ g)(x)$ if $f(x) = \frac{1}{x-5}$ and $g(x) = \sqrt{x}$.

Step 1: Identify the domain of the inner function $g(x)$. Since it's $\sqrt{x}$, $x \geq 0$.

Step 2: Identify the restriction of the outer function $f(g(x))$. The denominator $g(x) - 5$ cannot be zero.

Step 3: Solve the restriction: $\sqrt{x} - 5 \neq 0 \Rightarrow \sqrt{x} \neq 5 \Rightarrow x \neq 25$.

Final Answer: The domain is $[0, 25) \cup (25, \infty)$.

Type 4: Decomposition (The "Reverse" Process)

Question: Express $H(x) = (2x^2 - 1)^5$ as a composition of two functions $f$ and $g$ such that $H(x) = f(g(x))$.

Step 1: Identify the "inner" operation. Here, $2x^2 - 1$ is happening first. So, $g(x) = 2x^2 - 1$.

Step 2: Identify the "outer" operation applied to that result. The result is being raised to the power of $5$. So, $f(x) = x^5$.

Final Answer: $f(x) = x^5$ and $g(x) = 2x^2 - 1$. (This is vital for Calculus/Chain Rule later).

Type 5: Composition with Self (Iterative Functions)

Question: If $f(x) = \frac{x}{x+1}$, find $(f \circ f)(x)$.

Step 1: Set up the substitution: $f(f(x)) = \frac{f(x)}{f(x) + 1}$.

Step 2: Plug in the expression: $\dfrac{\frac{x}{x+1}}{\frac{x}{x+1} + 1}$.

Step 3: Multiply numerator and denominator by $(x+1)$ to clear the fractions: $\frac{x}{x + (x+1)}$.

Final Answer: $\frac{x}{2x+1}$.

6. More Worked Examples

Example 11: Finding the Domain

Find the domain of $f(x) = \sqrt{x - 5}$.

Solution: You can't take the square root of a negative.

So $x - 5 \geq 0$, which means $x \geq 5$.

Example 12: Evaluating Functions

If $f(x) = 3x^2 + 2x$, find $f(2)$.

Solution: $3(2)^2 + 2(2) = 3(4) + 4 = 12 + 4 = 16$.

Example 13: Piecewise Functions

A function that changes rules: $f(x) = x$ if $x > 0$, and $f(x) = -x$ if $x \leq 0$.

Solution: If $x = 2$, use first rule $\rightarrow 2$. If $x = -2$, use the second rule $\rightarrow -(-2) = 2$.

Example 14: Inverse Functions

Find the inverse of $f(x) = 2x + 1$.

Solution: Swap $x$ and $y$: $x = 2y + 1 \rightarrow x - 1 = 2y \rightarrow y = \frac{x-1}{2}$.

Example 15: Identifying Even/Odd

Is $f(x) = x^4$ even or odd?

Solution: $f(-x) = (-x)^4 = x^4$. Since $f(-x) = f(x)$, it is Even.

Example: Evaluating a Function

Question: Given $h(p) = p^2 + 2p$, find $h(4)$.

Solution: Substitute $4$ for every $p$ in the equation: $h(4) = (4)^2 + 2(4) = 16 + 8 = 24$.

Example: Finding Domain (Rational Function)

Question: Determine the domain of $f(x) = \frac{x-2}{x}$.

Solution: A rational function is undefined if the denominator is zero. Here, $x \neq 0$. Thus, the domain is all real numbers except $0$.

Example: Finding Domain (Square Root)

Question: Determine the domain of $f(x) = \sqrt{2 - x}$.

Solution: For the output to be a real number, the value under the square root must be $\geq 0$. Solve $2 - x \geq 0$, which gives $x \leq 2$. Domain: $(-\infty, 2]$.

Example: Function Composition

Question: If $f(x) = x^2$ and $g(x) = x + 3$, find $(f \circ g)(x)$.

Solution: Substitute $g(x)$ into $f$: $f(g(x)) = (x+3)^2 = x^2 + 6x + 9$.

Example: Operations on Functions

Question: If $f(x) = x + 2$ and $g(x) = 3x$, find $(f \cdot g)(x)$.

Solution: Multiply the functions: $(x+2)(3x) = 3x^2 + 6x$.

Example: Testing for Even Functions

Question: Determine if $f(x) = x^2$ is an even function.

Solution: Check if $f(-x) = f(x)$. $(-x)^2 = x^2$. Since $f(-x) = f(x)$, it is even and symmetric about the y-axis.

Example: Testing for Odd Functions

Question: Determine if $f(x) = x^3$ is an odd function.

Solution: Check if $f(-x) = -f(x)$. $(-x)^3 = -x^3$. Since $f(-x) = -f(x)$, it is odd and symmetric about the origin.

Example: Piecewise Functions

Question: Evaluate $f(3)$ if $f(x) = \{x^2 \text{ if } x < 2;\ 2x \text{ if } x \geq 2\}$.

Solution: Since $3 \geq 2$, use the second rule: $f(3) = 2(3) = 6$.

Example: Absolute Value Functions

Question: Solve $f(x) = |x - 5|$ for $f(2)$.

Solution: Substitute $x = 2$: $|2 - 5| = |-3| = 3$. This measures distance from $5$.

Example: Quadratic Function (Vertex Form)

Question: Find the vertex of $f(x) = -3x^2 + 6x - 1$.

Solution: Factor $-3$ out of the first two terms: $-3(x^2 - 2x) - 1$.

Complete the square inside: $-3(x^2 - 2x + 1) - 1 + 3 = -3(x-1)^2 + 2$.

The vertex is $(1, 2)$.

Limits and Continuity of a Function

Imagine driving a car from Lagos to Ibadan. Your GPS tracks your position every second. If a bridge at a specific point is washed out, you cannot physically stand on that point. However, by looking at your position just meters before the bridge and just meters after it, you can accurately predict exactly where that missing point lies.

In mathematics, this is what we call finding a limit. In your upcoming modules on derivatives and integrals, you will discover that calculus is entirely built upon this single concept: analyzing how a mathematical expression behaves as it gets closer and closer to a specific value, even if that value is impossible to plug in directly.

Foundational Concept

A limit is the value that a function approaches as the input variable approaches a specific number. We do not care what the function actually equals at that exact input; we only care about its behavior near it.

We write this mathematically as:

$\lim_{x \to c} f(x) = L$

This is read as: "The limit of $f(x)$ as $x$ approaches $c$ equals $L$."

Let us look at a numerical example. Consider the function:

$f(x) = \frac{x^2 - 4}{x - 2}$

If you try to calculate $f(2)$ directly, you get a mathematical disaster:

$f(2) = \frac{2^2 - 4}{2 - 2} = \frac{0}{0}$

Division by zero is undefined.

However, let us build a table of values to see what happens as $x$ gets extremely close to $2$ from both sides:

$x$ from the left	$f(x)$	$x$ from the right	$f(x)$
1.9	3.9	2.1	4.1
1.99	3.99	2.01	4.01
1.999	3.999	2.001	4.001
1.9999	3.9999	2.0001	4.0001

Notice how as $x$ crawls closer to $2$, the output $f(x)$ crawls closer to $4$. Therefore, even though $f(2)$ does not exist, $\lim_{x \to 2} f(x) = 4$.

The Laws of Limits

Evaluating limits from first principles every time would be exhausting. Fortunately, mathematicians have established a set of laws that allow us to break complicated limits into simpler parts. These laws work provided the individual limits involved actually exist.

Suppose $\lim f(x) = L$ and $\lim g(x) = M$ as $x \to a$, and let $c$ be any constant. Then:

Law	Statement
Law 1 — Constant Rule	$\lim c = c$
Law 2 — Identity Rule	$\lim x = a$
Law 3 — Sum Rule	$\lim [f(x) + g(x)] = L + M$
Law 4 — Difference Rule	$\lim [f(x) - g(x)] = L - M$
Law 5 — Constant Multiple Rule	$\lim [c \cdot f(x)] = c \cdot L$
Law 6 — Product Rule	$\lim [f(x) \cdot g(x)] = L \cdot M$
Law 7 — Quotient Rule	$\lim [f(x) / g(x)] = L / M$, provided $M \neq 0$
Law 8 — Power Rule	$\lim [f(x)]^n = L^n$
Law 9 — Root Rule	$\lim \sqrt[n]{f(x)} = \sqrt[n]{L}$, provided $L > 0$ when $n$ is even

Example 2.2 — Applying the Limit Laws

Evaluate: $\lim (3x^2 - 5x + 7)$ as $x \to 2$.

We apply the limit laws step by step, naming each law as we go.

Apply the Sum/Difference Rule (Laws 3 & 4):

$= \lim(3x^2) - \lim(5x) + \lim(7)$

Apply the Constant Multiple Rule (Law 5):

$= 3 \cdot \lim(x^2) - 5 \cdot \lim(x) + \lim(7)$

Apply the Power Rule (Law 8) and Identity Rule (Law 2) and Constant Rule (Law 1):

$= 3 \cdot (2^2) - 5 \cdot (2) + 7$

Arithmetic:

$= 3 \cdot 4 - 10 + 7 = 12 - 10 + 7 = 9$

Answer: 9

Observation: For a polynomial function $f(x)$, $\lim f(x)$ as $x \to a$ is always equal to $f(a)$. This is because polynomials are continuous everywhere. This means that for polynomials, you can simply substitute $x = a$ directly.

Detailed Breakdown

Step 1: Evaluate by Direct Substitution

Whenever you are asked to find a limit, your very first action should always be to plug the target number directly into the function. If the function is a simple polynomial or rational function and does not result in a zero in the denominator, this direct value is your limit.

Step 2: Resolve Indeterminate Forms ($\frac{0}{0}$)

If direct substitution results in $\frac{0}{0}$, the limit is not necessarily non-existent. It simply means the true answer is hidden. You must use algebra to simplify the expression and eliminate the terms causing the zero. The two most common techniques for this are:

Factorization: Breaking polynomials into products to cancel out the problematic terms.

Rationalization: Multiplying the numerator and denominator by a conjugate expression to eliminate square roots.

Step 3: Check One-Sided Limits

For a limit to exist globally, the function must approach the exact same value from both sides.

Left-hand limit ($\lim_{x \to c^-} f(x)$): The value $f(x)$ approaches as $x$ gets closer to $c$ using numbers smaller than $c$.

Right-hand limit ($\lim_{x \to c^+} f(x)$): The value $f(x)$ approaches as $x$ gets closer to $c$ using numbers larger than $c$.

The fundamental rule of limits states:

$\lim_{x \to c} f(x) = L \iff \lim_{x \to c^-} f(x) = L \text{ and } \lim_{x \to c^+} f(x) = L$

If the left-hand limit does not equal the right-hand limit, the general limit does not exist (DNE).

Example 1: Factorization Method

Question: Evaluate the limit:

$\lim_{x \to 3} \frac{x^2 - 9}{x - 3}$

Solution:

Try Direct Substitution: Plug in $x = 3$.

$\frac{3^2 - 9}{3 - 3} = \frac{0}{0}$

This is an indeterminate form. We must alter the function algebraically.

Factor the Numerator: Recognize that $x^2 - 9$ is a difference of two squares: $x^2 - 3^2 = (x-3)(x+3)$.

$\lim_{x \to 3} \frac{(x-3)(x+3)}{x-3}$

Cancel Common Terms: Because we are evaluating a limit, $x$ approaches $3$ but is never exactly $3$. This means $(x-3) \neq 0$, allowing us to safely divide it out:

$\lim_{x \to 3} (x + 3)$

Apply Direct Substitution Again: Now substitute $x = 3$ into the simplified expression: $3 + 3 = 6$.

Therefore, $\lim_{x \to 3} \dfrac{x^2 - 9}{x - 3} = 6$.

Example 2: Rationalization Method

Question: Evaluate the limit:

$\lim_{x \to 0} \frac{\sqrt{x+1} - 1}{x}$

Solution:

Try Direct Substitution: Plug in $x = 0$. $\dfrac{\sqrt{0+1} - 1}{0} = \dfrac{1 - 1}{0} = \dfrac{0}{0}$.

We must clear the square root from the numerator using a conjugate.

Multiply by the Conjugate: The conjugate of $\sqrt{x+1} - 1$ is $\sqrt{x+1} + 1$.

Multiply both the top and bottom by this term:

$\lim_{x \to 0} \frac{(\sqrt{x+1} - 1)(\sqrt{x+1} + 1)}{x(\sqrt{x+1} + 1)}$

Expand the Numerator: Use the identity $(a-b)(a+b) = a^2 - b^2$:

$(\sqrt{x+1})^2 - (1)^2 = x + 1 - 1 = x$

Now substitute this back into our limit:

$\lim_{x \to 0} \frac{x}{x(\sqrt{x+1} + 1)}$

Cancel Common Terms: Divide the $x$ out of the numerator and denominator:

$\lim_{x \to 0} \frac{1}{\sqrt{x+1} + 1}$

Apply Direct Substitution: Plug in $x = 0$:

$\frac{1}{\sqrt{0+1} + 1} = \frac{1}{1+1} = \frac{1}{2}$

Therefore, $\lim_{x \to 0} \dfrac{\sqrt{x+1} - 1}{x} = \dfrac{1}{2}$.

Using Expansion Method for Evaluating Limit

The core idea of the expansion method is that near $x = 0$, complex functions can be perfectly approximated by a custom polynomial.

For instance, if you graph $f(x) = \sin(x)$ and the simple line $f(x) = x$ near the origin, they look almost identical. If you want more accuracy, you add more terms:

$\sin(x) \approx x - \frac{x^3}{3!} + \frac{x^5}{5!}$

Here are the four standard Maclaurin series expansions you must memorize for your first-year exams. They are your primary tools for this method:

$e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots$

$\sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots$

$\cos(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots$

$\ln(1 + x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots$

(Note: The exclamation point denotes a factorial. For example, $3! = 3 \times 2 \times 1 = 6$, and $2! = 2 \times 1 = 2$.)

Detailed Breakdown

Step 1: Identify the Target Condition

The expansion method is most powerful when you are evaluating a limit where $x$ approaches $0$ ($\lim_{x \to 0}$) and direct substitution yields $\frac{0}{0}$.

Step 2: Substitute the Series Expansions

Replace the trigonometric or exponential functions in your limit with their corresponding polynomial series. Usually, writing out the first two or three terms of the series is more than enough to solve the problem.

Step 3: Group and Simplify the Terms

Group like powers of $x$ together in both the numerator and the denominator.

Step 4: Factor and Cancel the Lowest Power of $x$

The term causing the $\frac{0}{0}$ problem is always the lowest power of $x$ present in the expression. Factor this power out from both the top and bottom, then cancel them out completely.

Step 5: Evaluate by Direct Substitution

Plug $x = 0$ into the remaining simplified expression. Every term that still contains an $x$ will become zero, leaving you with your final numerical answer.

Worked Examples

Example 1: Solving a Trigonometric Limit

Question: Evaluate the limit using the expansion method:

$\lim_{x \to 0} \frac{x - \sin(x)}{x^3}$

Solution:

Try Direct Substitution:

$\frac{0 - \sin(0)}{0^3} = \frac{0 - 0}{0} = \frac{0}{0}$

This is indeterminate. Since there is a $\sin(x)$ term, we will use its series expansion.

Substitute the Expansion for $\sin(x)$:

Recall that $\sin(x) = x - \dfrac{x^3}{3!} + \dfrac{x^5}{5!} - \cdots$. Let us plug this directly into our limit expression:

$\lim_{x \to 0} \frac{x - \left(x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots\right)}{x^3}$

Distribute the Negative Sign and Simplify the Numerator:

Carefully distribute the minus sign into the parentheses:

$\lim_{x \to 0} \frac{x - x + \frac{x^3}{3!} - \frac{x^5}{5!} + \cdots}{x^3}$

Notice that the terms $x$ and $-x$ cancel each other out:

$\lim_{x \to 0} \frac{\frac{x^3}{3!} - \frac{x^5}{5!} + \cdots}{x^3}$

Factor out and Cancel $x^3$:

Divide every term in the numerator by the denominator's $x^3$:

$\lim_{x \to 0} \left(\frac{x^3}{3!x^3} - \frac{x^5}{5!x^3} + \cdots\right) = \lim_{x \to 0} \left(\frac{1}{3!} - \frac{x^2}{5!} + \cdots\right)$

Apply Direct Substitution ($x = 0$):

Every higher-power term containing $x$ will vanish to zero:

$\frac{1}{3!} - 0 = \frac{1}{3!} = \frac{1}{3 \times 2 \times 1} = \frac{1}{6}$

Therefore, $\lim_{x \to 0} \dfrac{x - \sin(x)}{x^3} = \dfrac{1}{6}$.

Example 2: Mixing Exponential and Trigonometric Functions

Question: Evaluate the limit: $\lim_{x \to 0} \dfrac{e^x - 1 - x}{\cos(x) - 1}$.

Solution:

Try Direct Substitution:

$\frac{e^0 - 1 - 0}{\cos(0) - 1} = \frac{1 - 1 - 0}{1 - 1} = \frac{0}{0}$

We must expand both $e^x$ in the numerator and $\cos(x)$ in the denominator.

Substitute Both Series Expansions:

$e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots$

$\cos(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots$

Substitute these into the limit:

$\lim_{x \to 0} \frac{\left(1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots\right) - 1 - x}{\left(1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots\right) - 1}$

Simplify the Numerator and Denominator:

In the numerator, $1 - 1 = 0$ and $x - x = 0$. In the denominator, $1 - 1 = 0$. Our expression collapses beautifully to:

$\lim_{x \to 0} \frac{\frac{x^2}{2!} + \frac{x^3}{3!} + \cdots}{-\frac{x^2}{2!} + \frac{x^4}{4!} - \cdots}$

Factor out and Cancel the Lowest Common Power ($x^2$):

Factor out $x^2$ from both the top and the bottom expressions:

$\lim_{x \to 0} \frac{x^2\left(\frac{1}{2!} + \frac{x}{3!} + \cdots\right)}{x^2\left(-\frac{1}{2!} + \frac{x^2}{4!} - \cdots\right)}$

Cancel the $x^2$ terms:

$\lim_{x \to 0} \frac{\frac{1}{2!} + \frac{x}{3!} + \cdots}{-\frac{1}{2!} + \frac{x^2}{4!} - \cdots}$

Apply Direct Substitution ($x = 0$):

$\frac{\frac{1}{2!} + 0}{-\frac{1}{2!} + 0} = \frac{\frac{1}{2}}{-\frac{1}{2}} = -1$

Therefore, $\lim_{x \to 0} \dfrac{e^x - 1 - x}{\cos(x) - 1} = -1$.

Common Pitfalls

1. The Variable Substitution Mistake: If your function contains a modified term like $\sin(2x)$, you cannot just use the expansion for $\sin(x)$. You must replace every single $x$ in the standard expansion formula with $(2x)$.

   For example: $\sin(2x) = (2x) - \dfrac{(2x)^3}{3!} + \cdots = 2x - \dfrac{8x^3}{6} + \cdots$

2. Truncating Too Early: If you only write down the very first term of a series (e.g., replacing $\cos(x)$ with just $1$), you might accidentally cancel everything out and get a false $0$ or $\frac{0}{0}$ again. Always write out at least up to the term matching the highest power in your denominator.

Common Pitfalls

1. The $\frac{0}{0} = 0$ or Undefined Error: Do not stop your work if you get $\frac{0}{0}$. It is a signal to do more algebra, not a final answer.
2. Forgetting the Limit Notation: Students often stop writing the symbol "$\lim_{x \to c}$" during intermediate steps. You must keep writing the limit operator until the exact moment you substitute the number into the variable.

Key Takeaways

1. Limit Concept: A limit tracks what value a function approaches, not what it explicitly equals at a point.
2. Indeterminate Form: Finding $\frac{0}{0}$ means you must simplify the function using factoring or rationalization.
3. Existence Condition: A general limit only exists if the left-sided limit matches the right-sided limit.

Continuity of a Function

Imagine drawing a graph on a piece of paper. If you can trace the entire curve from left to right without ever lifting your pen from the page, your function is continuous. If your pen has to jump over a missing point, skip across a gap, or shoot off to infinity, the function is discontinuous.

In real-world engineering, you want physical properties like the temperature of a solar panel or the voltage across a circuit to be continuous. Sudden breaks or jumps usually signal structural failures or explosive system changes.

Foundational Concept

A function $f(x)$ is continuous at a specific point $x = c$ if there are no breaks, holes, or jumps at that point. To prove this mathematically, the function must clear three strict hurdles.

A function $f(x)$ is continuous at $x = c$ if and only if:

1. $f(c)$ is defined (the point actually exists).
2. $\lim_{x \to c} f(x)$ exists (the left and right sides point to the same location).
3. $\lim_{x \to c} f(x) = f(c)$ (the missing hole is perfectly filled by the point).

If any single one of these three conditions fails, the function is discontinuous at $x = c$.

Detailed Breakdown

Step 1: Check if the Point Exists

Look at the function description. Can you plug $c$ into the function without breaking a mathematical law (like dividing by zero)? If you get an undefined result, stop immediately; the function is discontinuous.

Step 2: Verify the Limit Exists

Calculate the left-hand limit and the right-hand limit. If you are dealing with a piecewise function (a function that changes formulas at a certain boundary), evaluate the specific formula used for values smaller than $c$, then evaluate the formula used for values larger than $c$. If their values match, your limit exists.

Step 3: Compare Point and Limit

Check if your numeric answer from Step 1 matches your numeric answer from Step 2. If they are identical, the curve is unbroken, and the function is continuous.

Worked Examples

Example 1: Verifying Continuity of a Piecewise Function

Question: Determine if the following function is continuous at $x = 2$:

$f(x) = \begin{cases} x^2 + 1 & \text{if } x < 2 \\ 5 & \text{if } x = 2 \\ 3x - 1 & \text{if } x > 2 \end{cases}$

Solution:

We must test all three conditions of continuity at $x = 2$.

Condition 1: Does $f(2)$ exist?

Looking at the piecewise rules, when $x = 2$, the function explicitly says $f(x) = 5$. So $f(2) = 5$ (Condition 1 passes).

Condition 2: Does $\lim_{x \to 2} f(x)$ exist?

Because the formula changes at $x = 2$, we must check both one-sided limits.

Left-Hand Limit ($\lim_{x \to 2^-}$): Use the rule for $x < 2$, which is $x^2 + 1$.

$\lim_{x \to 2^-} (x^2 + 1) = 2^2 + 1 = 5$

Right-Hand Limit ($\lim_{x \to 2^+}$): Use the rule for $x > 2$, which is $3x - 1$.

$\lim_{x \to 2^+} (3x - 1) = 3(2) - 1 = 5$

Since both the left-hand limit and right-hand limit equal $5$, the general limit exists:

$\lim_{x \to 2} f(x) = 5 \text{ (Condition 2 passes)}$

Condition 3: Does the limit equal the point value?

Compare our answers: $\lim_{x \to 2} f(x) = 5$ and $f(2) = 5$.

$\lim_{x \to 2} f(x) = f(2) \text{ (Condition 3 passes)}$

Since all three conditions are satisfied, the function is continuous at $x = 2$.

Example 2: Finding a Missing Parameter for Continuity

Question: Find the value of the constant $k$ that makes the function continuous everywhere:

$f(x) = \begin{cases} kx + 3 & \text{if } x \leq 3 \\ 2x^2 - 1 & \text{if } x > 3 \end{cases}$

Solution: For the function to be continuous everywhere, it must specifically be continuous at the transition boundary $x = 3$. This means the left-hand limit must equal the right-hand limit at $x = 3$.

1. Find the Left-Hand Limit: Use the expression for $x \leq 3$.

$\lim_{x \to 3^-} (kx + 3) = k(3) + 3 = 3k + 3$

2. Find the Right-Hand Limit: Use the expression for $x > 3$.

$\lim_{x \to 3^+} (2x^2 - 1) = 2(3)^2 - 1 = 2(9) - 1 = 17$

3. Set Them Equal to Solve for $k$:

$3k + 3 = 17$

Subtract $3$ from both sides:

$3k = 14$

Divide by $3$:

$k = \frac{14}{3}$

Therefore, setting $k = \dfrac{14}{3}$ forces the two pieces of the graph to meet seamlessly at $x = 3$, making the function continuous.

Common Pitfalls

1. Only Checking One Side: When dealing with piecewise functions, students often evaluate only one of the expressions to find a limit. You must evaluate both sides of the boundary value separately.
2. Assuming Direct Value Rules Everything: A function can have a valid limit at a point while still being discontinuous if the actual point value is plotted elsewhere (a removable discontinuity). Always check condition 3!

Key Takeaways

1. Three-Part Test: Continuity requires a point to exist, a limit to exist, and for both values to match exactly.
2. Piecewise Transitions: To fix or verify continuity in split functions, match the left-sided expression to the right-sided expression at the split point.

Practice Exercises

1. If $f(x) = 5x - 7$, find $f(3)$.
2. What is the domain of $f(x) = \dfrac{1}{x-2}$?
3. Calculate $f(g(2))$ if $f(x) = x^2$ and $g(x) = x + 1$.
4. Determine if $f(x) = x^3$ is Even or Odd.
5. Find the inverse of $f(x) = x + 5$.
6. Solve $f(x) = 0$ for $f(x) = 2x - 10$.
7. Find $f(a + h)$ for $f(x) = 2x$.
8. If $f(x) = |x - 3|$, find $f(1)$.