Confirming Continuity
over an Interval
Learn to verify that a function has no breaks, jumps, or holes across an entire interval: the foundation of the Intermediate Value Theorem and integral calculus.
AP Calculus AB
What Is Continuity at a Point?
A function \( f \) is continuous at a point \( x = c \) when three conditions all hold simultaneously. If even one fails, the function is discontinuous at that point.
1
\( f(c) \) exists: the function is defined at \( c \).
2
\( \displaystyle\lim_{x \to c} f(x) \) exists: the left-hand and right-hand limits agree.
3
\( \displaystyle\lim_{x \to c} f(x) = f(c) \): the limit equals the function value.
Visualizing a Continuous Function
Why Interval Continuity Matters
Many of the most powerful theorems in calculus require that a function be continuous on a closed interval. Without confirming continuity first, you cannot apply them.
IVT
The Intermediate Value Theorem guarantees every output between \( f(a) \) and \( f(b) \) is achieved.
EVT
The Extreme Value Theorem guarantees a continuous function attains an absolute max and min on \[a,b]\].
FTC
The Fundamental Theorem of Calculus requires continuity on the integration interval \[a,b]\].
The Formal Definition
1
\( f \) is continuous at every point \( c \) in the open interval \((a,b)\).
2
\( f \) is right-continuous at \( a \): \(\displaystyle\lim_{x \to a^+} f(x) = f(a)\).
3
\( f \) is left-continuous at \( b \): \(\displaystyle\lim_{x \to b^-} f(x) = f(b)\).
One-Sided Limits at Endpoints
Open vs. Closed Interval Continuity
| Open \((a,b)\) | Closed \[a,b]\] | |
|---|---|---|
| Endpoints included? | No | Yes |
| Limit check at endpoints | Two-sided (interior only) | One-sided at \(a\) and \(b\) |
| \(f(a)\) and \(f(b)\) must exist? | Not required | Required |
| IVT / EVT applicable? | No | Yes |
Functions Known to Be Continuous
Several families of functions are continuous on their entire domain. You can cite these facts without proving continuity from the definition.
Polynomials
Continuous on \((-\infty,\infty)\). Always safe on any interval.
Trig Functions
\(\sin x\) and \(\cos x\) are continuous everywhere. \(\tan x\) is continuous where defined.
Exponentials
\(e^x\), \(a^x\) are continuous on \((-\infty,\infty)\).
How to Confirm Continuity
1
Identify the interval: Is it open \((a,b)\), closed \[a,b]\], or half-open?
2
Check the domain: Is \(f\) defined at every point in the interval (including endpoints for closed)?
3
Check interior continuity: Is \(f\) built from known continuous functions via valid operations (sum, product, composition, etc.)?
4
Check piecewise boundaries: If \(f\) changes formula, verify \(\displaystyle\lim_{x\to c^-}f(x)=\lim_{x\to c^+}f(x)=f(c)\) at each break.
5
Check endpoints (closed interval): Confirm the appropriate one-sided limit equals the function value.
Example 1: Polynomial
Confirm that \( f(x) = x^3 - 4x + 2 \) is continuous on \([-2, 3]\).
1
Domain: \(f\) is a polynomial, defined for all real numbers. Every point in \([-2,3]\) is in the domain. ✓
2
Interior: Polynomials are continuous on \((-\infty,\infty)\), so \(f\) is continuous on \((-2,3)\). ✓
3
Endpoints: Since polynomials are continuous everywhere, the one-sided limits at \(-2\) and \(3\) automatically match \(f(-2)\) and \(f(3)\). ✓
Example 2: Rational Function
Is \(\displaystyle g(x) = \frac{x+1}{x-2}\) continuous on \([0, 1]\)?
1
Domain issue: The denominator is zero when \(x = 2\). Does \(x=2\) lie in \([0,1]\)? No: \(2 \notin [0,1]\). ✓
2
Interior: The numerator and denominator are both polynomials (continuous everywhere). Their quotient is continuous wherever the denominator \(\neq 0\). On \((0,1)\), the denominator is never zero. ✓
3
Endpoints: \(g(0) = \frac{1}{-2} = -\frac{1}{2}\) and \(g(1) = \frac{2}{-1} = -2\). Both exist and match the one-sided limits. ✓
Example 3: Piecewise Function
Is the following function continuous on \([0, 4]\)?
1
Each piece: \(x^2\) and \(4x-4\) are polynomials, so each is continuous on its own sub-interval. ✓
2
At the break point \(x=2\): Left limit: \(\displaystyle\lim_{x\to 2^-} x^2 = 4\). Right limit: \(\displaystyle\lim_{x\to 2^+}(4x-4) = 4\). Function value: \(h(2) = 2^2 = 4\). All three equal 4. ✓
3
Endpoints: \(h(0) = 0\), \(h(4) = 12\). One-sided limits match. ✓
Visualizing the Piecewise Join
Types of Discontinuity
Removable
A "hole": the limit exists but does not equal \(f(c)\), or \(f(c)\) is undefined. Example: \(\frac{x^2-1}{x-1}\) at \(x=1\).
Jump
Left and right limits exist but are different. Common in piecewise functions where the pieces do not connect.
Infinite
A vertical asymptote: the function grows without bound as \(x \to c\). Example: \(\frac{1}{x}\) at \(x = 0\).
Common Mistakes
✕ Ignoring the Denominator
Forgetting to check where the denominator of a rational function equals zero inside the interval.
✕ Missing Piecewise Checks
Assuming two pieces connect without computing left and right limits at the boundary.
✕ Wrong Bracket Type
Confusing open \((a,b)\) with closed \[a,b]\] and forgetting to verify endpoint continuity for the closed case.
Summary Checklist
| Step | What to Check | Pass Condition |
|---|---|---|
| 1. Domain | \(f\) is defined on \[a,b]\] | No division by 0, no \(\sqrt{\text{neg}}\) |
| 2. Interior | Continuous on \((a,b)\) | Built from known continuous functions |
| 3. Break points | Left = Right = \(f(c)\) | All three values match at every break |
| 4. Left endpoint | \(\displaystyle\lim_{x\to a^+} f(x)\) | Equals \(f(a)\) |
| 5. Right endpoint | \(\displaystyle\lim_{x\to b^-} f(x)\) | Equals \(f(b)\) |
Question 1 of 5
Which of the following is NOT one of the three conditions for continuity at a point \(x = c\)?
Continuity requires that \(f(c)\) exists, the limit exists, and they are equal. The existence of the derivative \(f'(c)\) is the condition for differentiability, not continuity.
Question 2 of 5
Is \(\displaystyle f(x) = \frac{1}{x - 3}\) continuous on the interval \([0, 2]\)?
The only discontinuity is at \(x = 3\), which is not in \([0, 2]\). Since the denominator is never zero on this interval, \(f\) is continuous on \([0, 2]\).
Question 3 of 5
Consider the piecewise function: $$f(x) = \begin{cases} 2x + 1 & x < 3 \\ 10 & x = 3 \\ x^2 & x > 3 \end{cases}$$ Is \(f\) continuous at \(x = 3\)?
The left-hand limit is \(2(3)+1 = 7\) and the right-hand limit is \(3^2 = 9\). Since \(7 \neq 9\), the two-sided limit does not exist, so \(f\) is not continuous at \(x = 3\). The most fundamental reason is that the left and right limits disagree.
Question 4 of 5
For a function to be continuous on the closed interval \([a, b]\), which one-sided limit must hold at the left endpoint \(a\)?
At the left endpoint \(a\) of \[a,b]\], we approach from the right (from inside the interval), so we need \(\displaystyle\lim_{x \to a^+} f(x) = f(a)\). The left-hand limit is irrelevant since the function need not be defined to the left of \(a\).
Question 5 of 5
Let \(\displaystyle f(x) = \frac{x^2 - 9}{x - 3}\). On which interval is \(f\) continuous?
The function has a removable discontinuity at \(x = 3\) (the denominator is zero). The interval \([4, 7]\) does not contain \(x = 3\), so \(f\) is continuous there. Both \([0,5]\) and \([2,4]\) include \(x=3\), and \((-\infty,\infty)\) includes it as well.