Continuity

A function $f$ is **continuous at $x = a$** if its graph has no break there: you can draw through the point without lifting your pencil. In the language of limits, three conditions must all be true: 1. $f(a)$ is **defined** (there is a point on the graph at $x = a$) 2. $\lim_{x \to a} f(x)$ **exists** (the left-hand and right-hand limits are equal) 3. $\lim_{x \to a} f(x) = f(a)$ (the limit **equals** the function value) $$f \text{ is continuous at } a \iff \lim_{x \to a} f(x) = f(a)$$ If any one condition fails, $f$ has a **discontinuity** at $x = a$. The different kinds of breaks (holes, jumps, vertical asymptotes) are sorted out in the lesson Types of Discontinuity. **Continuity on an interval**: $f$ is continuous on an open interval $(a, b)$ if it is continuous at every point inside it. On a closed interval $[a, b]$ we also need $\lim_{x \to a^+} f(x) = f(a)$ and $\lim_{x \to b^-} f(x) = f(b)$ at the endpoints. **Functions that are continuous on their whole domain**: - polynomials (everywhere) - rational functions (everywhere except where the denominator is $0$) - exponential functions $b^x$ and $e^x$ (everywhere) - logarithms $\log_b x$ and $\ln x$ (for $x > 0$) - $\sqrt{x}$ (for $x \geq 0$) Sums, differences, products, quotients (where the denominator is not $0$) and compositions of continuous functions are continuous too. This is exactly why direct substitution works: for a continuous function, $\lim_{x \to a} f(x) = f(a)$.

Checking Continuity of a Polynomial

Is $f(x) = x^2 - 3x + 1$ continuous at $x = 2$? Check all three conditions.

Condition 1: is $f(2)$ defined?: $f(2) = 2^2 - 3 \cdot 2 + 1 = 4 - 6 + 1$ = $f(2) = -1$

Condition 2: does the limit exist?: $f$ is a polynomial, so the limit can be found by direct substitution: $\lim_{x \to 2} (x^2 - 3x + 1) = -1$ = The limit exists and equals $-1$

Condition 3: compare the limit and the value: $\lim_{x \to 2} f(x) = -1 = f(2)$ = All three conditions hold

Answer: Yes, $f$ is continuous at $x = 2$ (in fact, as a polynomial it is continuous everywhere).

When the Limit and the Value Disagree

Let $g(x) = \frac{x^2 - 9}{x - 3}$ for $x \neq 3$ and $g(3) = 5$. Is $g$ continuous at $x = 3$?

Condition 1: is $g(3)$ defined?: The definition gives $g(3) = 5$ directly = Yes, $g(3) = 5$

Condition 2: find the limit: $\frac{x^2 - 9}{x - 3} = \frac{(x - 3)(x + 3)}{x - 3} = x + 3$ for $x \neq 3$, so $\lim_{x \to 3} g(x) = 3 + 3$ = The limit exists and equals $6$

Condition 3: compare: $\lim_{x \to 3} g(x) = 6$, but $g(3) = 5$ = $6 \neq 5$: condition 3 fails

How could we repair it?: Redefine the single value $g(3) = 6$ = Then $g$ would be continuous at $3$

Answer: No. The limit exists ($6$) and $g(3)$ is defined ($5$), but they are not equal, so $g$ is not continuous at $x = 3$. Setting $g(3) = 6$ would make it continuous. The second graph below shows this: an open circle at $(3, 6)$ and the filled point $g(3) = 5$ below it.

Where Is a Rational Function Continuous?

On which intervals is $h(x) = \frac{x + 1}{x^2 - 4}$ continuous?

Recall the rule: A rational function is continuous everywhere except where its denominator is $0$ = Find the zeros of $x^2 - 4$

Solve denominator $= 0$: $x^2 - 4 = (x - 2)(x + 2) = 0$ = $x = 2$ or $x = -2$

Write the intervals: Remove $-2$ and $2$ from the number line = $(-\infty, -2) \cup (-2, 2) \cup (2, \infty)$

Answer: $h$ is continuous on $(-\infty, -2)$, on $(-2, 2)$ and on $(2, \infty)$. It is discontinuous only at $x = -2$ and $x = 2$, where it is undefined.

Finding k So a Piecewise Function Is Continuous

Find $k$ so that $f(x) = \begin{cases} x^2 + 1, & x \leq 2 \\ kx - 1, & x > 2 \end{cases}$ is continuous everywhere.

Each piece is a polynomial: Both pieces are continuous on their own intervals, so the only point to check is the breakpoint $x = 2$ = Check $x = 2$

Value and left-hand limit (first piece): $f(2) = 2^2 + 1$ and $\lim_{x \to 2^-} f(x) = 2^2 + 1$ = Both equal $5$

Right-hand limit (second piece): $\lim_{x \to 2^+} f(x) = k \cdot 2 - 1$ = $2k - 1$

Make the two sides meet: $2k - 1 = 5 \Rightarrow 2k = 6$ = $k = 3$

Check: With $k = 3$: $3 \cdot 2 - 1 = 5$, so both one-sided limits are $5 = f(2)$ = Continuous at $x = 2$

Answer: $k = 3$. Try it in the interactive graph below: move the slider to $k = 3$ and the two pieces join at the point $(2, 5)$.

Intermediate Value Theorem: Proving a Root Exists

Show that $x^3 + x - 1 = 0$ has a solution between $x = 0$ and $x = 1$.

Name the function and check continuity: $f(x) = x^3 + x - 1$ is a polynomial = $f$ is continuous on $[0, 1]$

Evaluate at the endpoints: $f(0) = 0 + 0 - 1$ and $f(1) = 1 + 1 - 1$ = $f(0) = -1$, $f(1) = 1$

Is $0$ between the two values?: $-1 < 0 < 1$ = Yes, the sign changes

Apply the IVT: A continuous function that goes from $-1$ to $1$ must take the value $0$ somewhere in between = There is a $c$ in $(0, 1)$ with $f(c) = 0$

Answer: By the Intermediate Value Theorem there is a number $c$ between $0$ and $1$ with $c^3 + c - 1 = 0$. The third graph below shows it: $c \approx 0.68$.

Continuity is the property that makes most of calculus work: - **Direct substitution** is only allowed because polynomials, exponentials and logarithms are continuous: the limit is just the function value. - **Root finding**: the Intermediate Value Theorem guarantees that a continuous function that changes sign must cross zero. Calculators and computers use this to locate solutions of equations. - **Modelling**: temperature, height, distance and the volume of water in a tank change continuously with time. Prices, tax brackets and parking fees often do not, and a designer has to choose constants so there is no sudden jump. - **Engineering**: two pieces of road, rail or pipe must meet at the same height, which is exactly the condition that a piecewise function is continuous at the joint. Knowing where a function is continuous tells you where you can trust its graph, where you can substitute, and where something dramatic (a hole, a jump, an asymptote) happens.

Mistake: Checking only that $f(a)$ exists and concluding that $f$ is continuous

Why: A function can have a value at $a$ and still jump there. For example $f(x) = 1$ for $x < 0$ and $f(x) = 2$ for $x \geq 0$ has $f(0) = 2$, but the left-hand limit is $1$.

Correct: Always check all three conditions: $f(a)$ is defined, the limit exists, and the limit equals $f(a)$.

Mistake: Thinking that if the limit exists, the function must be continuous

Why: The limit ignores what happens exactly at $x = a$. In Example 2 the limit is $6$, but the point sits at $g(3) = 5$.

Correct: After finding the limit, compare it with the actual value $f(a)$. They must be equal.

Mistake: Solving for $k$ by plugging the breakpoint into only one piece

Why: Continuity at a breakpoint is about the two pieces meeting, so one piece alone gives no equation for $k$.

Correct: Evaluate both pieces at the breakpoint (left-hand and right-hand limits) and set them equal: in Example 4, $2^2 + 1 = 2k - 1$.

Mistake: Using the Intermediate Value Theorem on a function that is not continuous on the interval

Why: $f(x) = \frac{1}{x}$ has $f(-1) = -1$ and $f(1) = 1$, but it is never $0$. The IVT does not apply because $f$ is undefined at $x = 0$.

Correct: Before using the IVT, confirm that $f$ is continuous on the whole closed interval $[a, b]$.

Fair Parking Fees

A parking garage uses one rate for short stays and another for long stays. If the formulas do not agree at the switch point, the price jumps and a driver who stays one minute longer pays much more or much less.

A fee of $C(t) = 5t$ dollars for up to 2 hours and $C(t) = 3t + 4$ dollars after that gives $C(2) = 10$ dollars from both formulas, so the fee has no jump.

Hiking Altitude and the IVT

Your altitude on a hike changes continuously with time: you cannot teleport from one height to another. The Intermediate Value Theorem then guarantees that you passed through every height between your start and your finish.

If you start at 800 m and reach a summit at 2000 m, at some moment you were at exactly 1500 m, even if you did not look at your watch then.

Joining Two Pieces of Road

Road and railway designers describe a track with different formulas on different sections. At every joint the heights must match, which is the condition for a piecewise function to be continuous.

A ramp with height $y = 0.5x$ for $x \leq 2$ followed by $y = -0.25x + 1.5$ for $x > 2$ (in meters) meets at $y = 1$ m, so there is no step.

$f$ is continuous at $x = a$ when $f(a)$ is defined, $\lim_{x \to a} f(x)$ exists, and $\lim_{x \to a} f(x) = f(a)$

If any of the three conditions fails, $f$ has a discontinuity at $a$ (the kinds are covered in Types of Discontinuity)

Polynomials, exponentials, $\sin$ and $\cos$ are continuous everywhere; rational functions, logarithms and roots are continuous on their domains

On a closed interval $[a, b]$, continuity at the endpoints uses one-sided limits

To make a piecewise function continuous, set the left-hand limit, the right-hand limit and the value at each breakpoint equal and solve for the constant

Intermediate Value Theorem: if $f$ is continuous on $[a, b]$ and $N$ is between $f(a)$ and $f(b)$, then $f(c) = N$ for some $c$ in $(a, b)$

Q: Is $f(x) = \frac{1}{x}$ a continuous function?

A: It is continuous on its domain, that is, at every $x \neq 0$. At $x = 0$ it is not defined, so it is not continuous there. When someone says a function is continuous without naming a point, they usually mean continuous on its whole domain.

Q: Why does direct substitution give the limit for polynomials?

A: Because polynomials are continuous everywhere, and continuity means exactly $\lim_{x \to a} f(x) = f(a)$. So the limit is the function value. If substitution gives $\frac{0}{0}$, the function is not continuous at that point, and you have to simplify first.

Q: Does the IVT tell me where the root is?

A: No. It only guarantees that at least one $c$ exists in $(a, b)$. To narrow it down, you can halve the interval again and again and keep the half where the sign changes (the bisection method).

Q: Can a function be continuous everywhere and still have a sharp corner?

A: Yes. $f(x) = |x|$ has a corner at $x = 0$, but you can draw it without lifting your pencil, and $\lim_{x \to 0} |x| = 0 = f(0)$. Corners matter for derivatives, not for continuity.

Exit Presentation

Continuity

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Pre-Calculus / Limits

Continuity

Learn the three-part definition of continuity, where common functions are continuous, how to find a constant that makes a piecewise function continuous, and the Intermediate Value Theorem.

Advanced30 min

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