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Calculus AB · Unit 1 · 1.13
AP Calculus · Limits and Continuity
Removing Discontinuities
You can fill a hole, but you cannot bridge a jump. The limit decides which one you have.
key idea$\text{Removable}\iff\lim_{x\to c}f\text{ exists. Then set }f(c)=\lim_{x\to c}f$
The procedure
You can only patch a hole. If the two-sided limit exists, define the function's value there to be that limit and the discontinuity vanishes. If the limit does not exist, no value of $f(c)$ can help — there is nothing for it to match.
1
Check the form at $c$
Substitute. A $\tfrac00$ result means a shared factor and a probable removable discontinuity. A $\tfrac{k}{0}$ result means an infinite discontinuity — stop, it cannot be removed.
in the function out removable or not
2
Compute the limit
Factor and cancel, rationalise, or take one-sided limits for a piecewise function. No new technique is needed here.
in the expression out the value $L$
3
Set the value equal to the limit
Define $f(c)=L$, or solve for the unknown constant that makes $f(c)=L$ true. For a piecewise function with two unknowns, you get one equation per seam.
in $L$ out the value or the constant
4
Verify all three conditions
Value exists, limit exists, and they are equal. State it — the verification is part of the answer, not an optional check.
out a justified conclusion
| Substituting gives | Removable? | What to do |
|---|
| $\tfrac00$ | usually yes | cancel, take the limit, set $f(c)=L$ |
| $\tfrac{k}{0}$, $k\neq0$ | no | infinite discontinuity — nothing to remove |
| different one-sided limits | no | jump — non-removable |
| a number, but $f(c)$ differs | yes | redefine $f(c)$ to that number |
| a number, and $f(c)$ matches | — | already continuous |
Only the two yes rows are repairable, and both are cases where the limit exists.
Worked examples
EX 1
Fill the hole
$\dfrac{x^{2}-9}{x-3}$Define $f(3)$ so that $f(x)=\dfrac{x^{2}-9}{x-3}$ becomes continuous at $x=3$.
1
Check the form
$\tfrac{0}{0}$ — a shared factor, so removable.
2
Compute the limit
$\dfrac{(x-3)(x+3)}{x-3}=x+3$ for $x\neq3$, so $\lim_{x\to3}f=6$.
3
Set the value
Define $f(3)=6$.
4
Verify
$f(3)=6$ exists; the limit is $6$; they are equal. Continuous.
EX 2
Solve for a constant
one seamFind $a$ so that $g(x)=\begin{cases}\dfrac{x^{2}-a^{2}}{x-a},& x\neq a\\ 10,& x=a\end{cases}$ is continuous, given $a\gt 0$.
1
Compute the limit in terms of $a$
$\dfrac{(x-a)(x+a)}{x-a}=x+a$, so $\lim_{x\to a}g=2a$.
2
Match the value
Continuity needs $g(a)=\lim_{x\to a}g$, i.e. $10=2a$.
4
Verify
With $a=5$: the limit is $10$ and $g(5)=10$. Continuous.
5
Note the structure
The unknown sat inside the function, not just at the point — so the limit itself depended on $a$.
EX 3
Two unknowns, two seams
systemFind $a$ and $b$ making $h(x)=\begin{cases}x+1,& x\lt 1\\ ax+b,& 1\le x\lt 3\\ 2x-5,& x\ge 3\end{cases}$ continuous everywhere.
1
Seam at $x=1$
Left limit $=2$; the middle piece at $1$ gives $a+b$. So $a+b=2$.
2
Seam at $x=3$
Middle piece approaches $3a+b$; right piece at $3$ gives $1$. So $3a+b=1$.
3
Solve the system
Subtracting: $2a=-1$, so $a=-\tfrac12$ and $b=\tfrac52$.
4
Verify both seams
At $1$: $-\tfrac12+\tfrac52=2$. ✓ At $3$: $-\tfrac32+\tfrac52=1$. ✓
5
Why one equation per seam
Each junction contributes exactly one continuity condition. Two unknowns need two seams — count before you start.
EX 4
When it cannot be removed
the limit caseCan $p(x)=\dfrac{x+2}{x-4}$ be made continuous at $x=4$ by defining $p(4)$?
1
Check the form
$\tfrac{6}{0}$ — not indeterminate, so nothing cancels.
2
Examine the one-sided limits
As $x\to4^{+}$, $p\to+\infty$; as $x\to4^{-}$, $p\to-\infty$.
3
Apply the criterion
The two-sided limit does not exist, so there is no value for $p(4)$ to equal.
4
Conclude
The discontinuity is non-removable (infinite). No definition of $p(4)$ makes $p$ continuous at $4$.
5
Contrast
In Example 1 the offending factor cancelled; here it does not. That single difference decides removability.
Where it goes wrong
Trying to patch a non-removable discontinuity
A jump or an infinite discontinuity cannot be fixed by any value. Check that the limit exists before solving for anything.
Only checking one side at a seam
Continuity needs both one-sided limits and the value to agree. Matching just the left one is the classic piecewise error.
Forgetting to verify
Solving for $k$ is the work; stating that all three conditions now hold is the justification. Questions ask for both.
Miscounting the unknowns
Each seam gives one equation. Two unknowns need two seams — if you have more unknowns than seams the system is underdetermined.