AP Calculus AB/BC · Unit 1 · Topic 1.13
Removing Discontinuities
Choose a point value or parameter that repairs an existing finite two-sided limit.
1. Topic Focus
Build the language of limits, connect numerical, graphical, and algebraic representations, and use continuity theorems with verified hypotheses.
This topic: Choose a point value or parameter that repairs an existing finite two-sided limit.
2. Key Relationship
Read every symbol with its domain, direction, units, and hypotheses before applying the relationship.
3. Visual Connection
4. Worked Example
Extend (x²-4)/(x-2) continuously by defining f(2)=4.
Write the governing relationship first, carry out the algebra cleanly, and finish with a sentence that answers the mathematical question.
5. Concept Development
Removal is possible exactly when a finite limit exists
Suppose \(f\) is not continuous at \(x=a\). Changing only the value at \(a\) can repair continuity if and only if the finite two-sided limit exists:
The continuous extension keeps every nearby value unchanged and assigns the missing point the approached value:
If \(f(a)\) already exists but differs from \(L\), replace it with \(L\). The repair value is unique because continuity requires equality with the one existing finite limit.
Diagnose before trying to repair
| Nearby behavior at \(a\) | Can one point value repair it? | Reason |
|---|---|---|
| Both sides approach the same finite \(L\) | Yes | Define or redefine \(f(a)=L\) |
| Finite one-sided limits are unequal | No | A point value cannot make the surrounding branches agree |
| At least one side is unbounded | No | A point cannot remove a vertical asymptote |
| Outputs oscillate without a finite limit | No | No single approached height exists to assign |
| Only one side exists at a domain endpoint | Sometimes | Use the appropriate one-sided continuity condition for that domain |
A complete repair workflow
- Locate the discontinuity from the graph, formula, table, or piecewise boundary.
- Compute both one-sided limits or otherwise establish the finite two-sided limit.
- If the limit is not one finite number, state that point redefinition is impossible.
- If the limit is \(L\), define the point value as \(f(a)=L\).
- Verify \(\lim\limits_{x\to a}f(x)=f(a)\) after the redefinition.
- If continuity is required over an interval, check all remaining points and endpoints too.
Rational functions: compare factor multiplicities
Suppose the numerator contains \((x-a)^m\) and the denominator contains \((x-a)^n\), with all other factors nonzero at \(a\).
| Multiplicity comparison | After cancellation | Behavior at \(a\) |
|---|---|---|
| \(m>n\) | A numerator factor remains | Removable with extension value 0 |
| \(m=n\) | No \(x-a\) factor remains | Removable with a finite nonzero value determined by remaining factors |
| \(m<n\) | A denominator factor remains | Usually a vertical asymptote; not removable by a point value |
Always preserve the original restriction \(x\ne a\) while computing the limit. Cancellation reveals the nearby rule; assigning the limit afterward fills the excluded point.
Other algebraic forms can hide removable points
- Radicals: multiply by a conjugate to reveal the finite nearby expression.
- Complex fractions: combine the smaller fractions or clear their common denominator.
- Trigonometric expressions: rewrite using identities and foundational limits in radians.
- Absolute values: check both one-sided formulas; a finite common value is required.
The method used to compute the limit may change, but the repair rule remains \(f(a)=\lim\limits_{x\to a}f(x)\).
Parameters can change the nearby structure
An unknown parameter may occur in a numerator, denominator, or piecewise branch. In that case, solve first for the parameter that creates a finite common limit. Then calculate the required point value.
For a rational expression with denominator zero at \(a\), a necessary first step for removability is often that the numerator also vanish:
This condition creates a possible common factor, but factoring and evaluating the simplified limit are still required.
Piecewise boundaries require three-way agreement
For a boundary \(x=a\), continuity requires
If a parameter changes an entire branch, solve for agreement between the branch limits. If a parameter changes only the point value, it can repair only an already removable discontinuity.
Repairing several holes
A function can have more than one removable discontinuity. Treat each excluded input independently: simplify the nearby rule, compute the limit at each hole, and assign a potentially different value at each point. The resulting extended function can then be continuous across a larger interval.
What a repair does and does not change
- It changes only the function value at the repaired input.
- It does not alter limits at that input because limits use nearby values.
- It does not remove unrelated discontinuities elsewhere.
- It does not turn a jump, asymptote, or oscillation into continuous behavior.
- It may enlarge the domain and merge adjacent continuity intervals through the repaired point.
Common errors
- Assigning \(f(a)=0\) merely because substitution produced \(0/0\).
- Finding the parameter that makes the numerator zero but never evaluating the resulting limit.
- Canceling terms instead of common factors.
- Choosing one of two unequal one-sided limits as the point value.
- Calling a remaining denominator factor removable.
- Forgetting to define values at every hole when continuity on a full interval is requested.
6. Detailed Worked Example and Error Check
Example 1: Fill a factored hole.
For nearby inputs, \(f(x)=x+2\). Therefore
The unique continuous extension defines \(\boxed{f(2)=4}\).
Example 2: Remove a radical discontinuity.
Multiply by the conjugate:
Thus \(\lim\limits_{x\to2}g(x)=1/6\), so define \(\boxed{g(2)=1/6}\).
Example 3: Choose a structural parameter, then fill the point. Let
For the discontinuity to be removable, the numerator must vanish at 2:
Then \(x^2+x-6=(x-2)(x+3)\), so the nearby rule is \(x+3\) and
Example 4: Match a piecewise boundary. Define
The right-hand limit and \(p(2)\) are 4. The left-hand limit is \(2c+1\), so
With this value, both branches and the point value agree at 2.
Example 5: Repair two holes independently.
Factoring gives \(r(x)=x\) on its original domain. Therefore
Defining \(\boxed{r(-1)=-1}\) and \(\boxed{r(1)=1}\) produces the continuous extension \(r(x)=x\) on all real numbers.
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.
Determine whether each discontinuity can be removed. When possible, give every required parameter and point value and verify continuity.
(a) Define \(f(5)\) so that \(f(x)=\frac{x^2-25}{x-5}\) for \(x\ne5\) becomes continuous.
(b) Define \(g(0)\) for \(g(x)=\frac{\sqrt{x+4}-2}{x}\), \(x\ne0\).
(c) Choose \(k\) and \(h(3)\) so that \(h(x)=\frac{x^2+kx-12}{x-3}\), \(x\ne3\), extends continuously.
(d) Define the missing value of \(p(x)=\frac{(x-2)^3}{(x-2)^2(x+1)}\) at \(x=2\).
(e) Decide whether any value at \(x=1\) can make \(q(x)=\frac{x-1}{(x-1)^2}\) continuous there.
(f) Find \(m\) so that \(F(x)=mx-2\) for \(x<3\) and \(F(x)=x+4\) for \(x\ge3\) is continuous at 3.
(g) Find both point values needed to extend \(R(x)=\frac{x^3-x}{x^2-1}\) continuously across its original denominator zeros.
(h) Explain why no choice of \(s(0)\) can make \(s(x)=|x|/x\) for \(x\ne0\) continuous at 0.
Check the solution
In part (a), the nearby rule is \(x+5\), so define \(f(5)=10\). In part (b), conjugate simplification gives \(1/(\sqrt{x+4}+2)\), so define \(g(0)=1/4\). In part (c), numerator cancellation requires \(9+3k-12=0\), so \(k=1\); the nearby rule becomes \(x+4\), so define \(h(3)=7\). In part (d), cancellation gives \((x-2)/(x+1)\), whose limit at 2 is 0, so define \(p(2)=0\). In part (e), one denominator factor remains and \(q(x)=1/(x-1)\) for \(x\ne1\); its one-sided limits are opposite infinities, so no point value works. In part (f), the left limit is \(3m-2\), while the right limit and \(F(3)\) are 7; hence \(m=3\). In part (g), the nearby rule is \(x\), so define \(R(-1)=-1\) and \(R(1)=1\). In part (h), the left-hand limit is \(-1\) and the right-hand limit is 1, so no finite two-sided limit exists and one assigned point cannot remove the jump.