AP Calculus AB/BC · Unit 1 · Topic 1.11
Defining Continuity at a Point
Verify the function value, existence of the limit, and equality between them.
1. Topic Focus
Build the language of limits, connect numerical, graphical, and algebraic representations, and use continuity theorems with verified hypotheses.
This topic: Verify the function value, existence of the limit, and equality between them.
2. Key Relationship
Read every symbol with its domain, direction, units, and hypotheses before applying the relationship.
3. Visual Connection
4. Worked Example
A piecewise boundary is continuous only when both side formulas and the assigned value agree.
Write the governing relationship first, carry out the algebra cleanly, and finish with a sentence that answers the mathematical question.
5. Concept Development
The definition has three necessary conditions
A function \(f\) is continuous at an interior point \(x=a\) exactly when all three statements are true:
| Condition | Mathematical statement | Question it answers |
|---|---|---|
| 1. Point value | \(f(a)\) is defined | Is there an assigned output at the target? |
| 2. Nearby agreement | \(\lim\limits_{x\to a}f(x)\) exists as a finite number | Do both nearby branches approach one height? |
| 3. Match | \(\lim\limits_{x\to a}f(x)=f(a)\) | Does the assigned point lie on the nearby trend? |
but the compact equation is valid only after both the finite limit and the point value are known to exist. It is a summary of all three conditions, not permission to skip them.
Why each condition is necessary
- If \(f(a)\) is undefined, the graph cannot include the point needed for continuity.
- If the two-sided limit does not exist, nearby values do not settle at one finite height, even if \(f(a)\) is defined.
- If the limit and function value differ, the nearby graph approaches one height while the assigned point sits at another.
No one condition implies the other two. A complete justification identifies every condition, or identifies the first failed condition and explains it precisely.
Expand the limit condition into one-sided statements
At an interior point, Condition 2 can be checked by comparing the one-sided limits:
Thus a fully expanded continuity statement is
This form is especially useful for piecewise functions because each side may use a different formula.
Read the three conditions in every representation
| Representation | Point value | Two-sided limit | Continuity conclusion |
|---|---|---|---|
| Graph | Height of the filled point at \(a\) | Common height approached by both branches | The filled point and approached height coincide |
| Table | Separate row at exactly \(a\) | Matching trends from below and above | The point row agrees with the common trend |
| Formula | Evaluate the assigned rule at \(a\) | Apply valid limit procedures to nearby rules | Compare the two results |
| Words | State the actual output | State the approached nearby output | Explain whether they are equal |
Direct substitution is a continuity argument
For a function already known to be continuous at \(a\), direct substitution works because
Polynomials are continuous at every real input. Rational functions are continuous where their denominators are nonzero. Root, logarithmic, and trigonometric expressions require their real-domain conditions. Do not use direct substitution as proof of continuity when it produces an undefined expression or violates the domain.
Piecewise continuity at a boundary
Suppose
If the pieces are continuous near \(a\), then continuity at the boundary requires
The expression containing the equality sign determines \(f(a)\), but both expressions determine the two one-sided limits. Substituting only into the equality piece checks the point value, not two-sided continuity.
Use continuity to choose parameters
When a piecewise function contains an unknown constant, translate continuity into an equation:
- Evaluate the left-hand limit.
- Evaluate the right-hand limit and the assigned point value.
- Set all required quantities equal.
- Solve for the parameter and substitute back to verify.
If the parameter affects only \(f(a)\), it can repair a removable discontinuity but cannot repair unequal or unbounded one-sided limits.
One-sided continuity at domain endpoints
At the left endpoint \(a\) of a domain, continuity relative to that domain uses
At a right endpoint \(b\), use \(\lim\limits_{x\to b^-}f(x)=f(b)\). A missing side outside the domain does not by itself create a discontinuity. Two-sided continuity remains the standard requirement at interior points.
Continuity of a composition at a point
If \(g(x)\to M\) as \(x\to a\) and \(f\) is continuous at \(M\), then
For the composition \(f\circ g\) itself to be continuous at \(a\), also require \(g\) to be continuous at \(a\), so that \(M=g(a)\). Check that \(g(a)\) lies in the domain where the outer function is continuous.
Common reasoning errors
- Saying “the graph has no break” without citing limit and function-value evidence.
- Checking only that \(f(a)\) exists.
- Checking only that both one-sided limits are finite without confirming they are equal.
- Using the formula selected at \(x=a\) for both sides of a piecewise limit.
- Writing \(\lim f(x)=f(a)\) when one side of the equation is undefined.
- Assuming continuity guarantees differentiability; a corner can be continuous without having a derivative.
- Requiring a nonexistent side at an endpoint of the stated domain.
6. Detailed Worked Example and Error Check
Example 1: Verify all three conditions. Let
At \(a=1\), the denominator is nonzero, so \(f(1)=2/3\) is defined. The numerator and denominator are continuous, and the quotient-law condition holds:
All three conditions hold, so \(f\) is \(\boxed{\text{continuous at }1}\).
Example 2: Condition 1 fails. For
the nearby expression simplifies to \(x+2\), so \(\lim\limits_{x\to2}g(x)=4\). However, \(g(2)\) is undefined. The limit exists, but the function is not continuous because Condition 1 fails.
Example 3: Condition 2 fails. Define
Here \(h(1)=4\) is defined, but
The two-sided limit does not exist, so \(h\) is not continuous at 1 even though the right-hand limit equals \(h(1)\).
Example 4: Condition 3 fails. Let \(p(x)=x^2\) for \(x\ne2\) and \(p(2)=7\). Then
The point value and finite limit both exist, but they do not match. Redefining \(p(2)=4\) would make the function continuous.
Example 5: Choose a parameter at a piecewise boundary. Let
The left-hand limit is \(6+k\). The right-hand limit and assigned value are both 8. Continuity requires
Substitution back into the left rule confirms that both sides and the point value equal 8.
7. AP Reasoning Routine
Read one-sided behavior first, choose a matching limit procedure, and justify conclusions with definitions or theorem conditions.
- Identify the representation and requested quantity.
- State the rule or theorem and verify its conditions.
- Keep exact values until the final requested approximation.
- Interpret sign, units, interval, and context.
Use the three-condition definition and state which condition fails whenever the function is not continuous.
(a) Determine whether \(f(x)=3x^2-2x+1\) is continuous at \(x=-1\).
(b) Determine whether \(g(x)=\frac{x+1}{x-2}\) is continuous at \(x=2\).
(c) Find \(c\) so that \(h(x)=cx+1\) for \(x<2\) and \(h(x)=x^2\) for \(x\ge2\) is continuous at 2.
(d) A graph has both branches approaching 6 at \(x=4\), but a filled point at \((4,-1)\). Test all three conditions.
(e) A table shows left outputs approaching 3, right outputs approaching 3, and \(p(5)=3\). State a complete continuity conclusion at 5 and identify what makes it an estimate if the table values are rounded.
(f) Suppose \(\lim\limits_{x\to a}q(x)=2\) and \(q(a)=2\). Is \(q\) continuous at \(a\)? Justify using the definition.
(g) Verify right continuity of \(r(x)=\sqrt{x-1}\) at the left endpoint \(x=1\) of its domain.
(h) If \(u\) is continuous at \(a\), \(u(a)=4\), and \(v\) is continuous at 4 with \(v(4)=-2\), determine \(\lim\limits_{x\to a}v(u(x))\) and explain why the composition is continuous at \(a\).
Check the solution
In part (a), polynomials are continuous everywhere; \(f(-1)=3+2+1=6\) and the limit is also 6. In part (b), \(g(2)\) is undefined and the one-sided behavior is unbounded, so Conditions 1 and 2 fail. In part (c), the left limit is \(2c+1\), while the right limit and \(h(2)\) are 4; continuity requires \(2c+1=4\), so \(c=3/2\). In part (d), the point value is defined and the finite two-sided limit exists, but \(6\ne-1\), so Condition 3 fails. In part (e), the common two-sided trend and point value are all 3, so the evidence supports continuity at 5; rounded finite table data provides numerical evidence rather than an exact analytical proof. In part (f), yes: the point value exists, the finite two-sided limit exists, and they are equal. In part (g), \(r(1)=0\) and \(\lim\limits_{x\to1^+}\sqrt{x-1}=0\), so \(r\) is right-continuous at its domain endpoint. In part (h), continuity of \(u\) gives \(u(x)\to4=u(a)\), and continuity of \(v\) at 4 gives \(v(u(x))\to v(4)=-2=v(u(a))\); hence the composition is continuous at \(a\).