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Calculus AB · Unit 1 · 1.11
AP Calculus · Limits and Continuity
Defining Continuity at a Point
One equation, three separate conditions — and a complete answer names all three.
key idea$f\text{ continuous at }c\iff \lim_{x\to c}f(x)=f(c)$
What it says
Continuity at a point
\[\lim_{x\to c}f(x)=f(c)\]
That one equation quietly demands three separate things, and a complete justification names all three. Checking only the last is the commonest way to lose the mark.
The three conditions
1. $f(c)$ exists — $c$ is in the domain. 2. $\lim_{x\to c}f(x)$ exists — both one-sided limits agree. 3. The two are equal. Fail any one and $f$ is discontinuous at $c$.
| Condition | What fails if it does not hold | Type of discontinuity |
|---|
| $f(c)$ exists | a hole where the graph is undefined | removable |
| $\lim_{x\to c}f$ exists | the two sides disagree, or blow up | jump / infinite |
| they are equal | a point plotted off the curve | removable |
Every discontinuity is one of these three conditions failing — the classification and the definition are the same content from two sides.
Reading the notation
Continuous at a pointThe three-condition test at a single $x=c$. This topic.
Continuous on an intervalThe point test holding at every interior point, with one-sided continuity at the endpoints. A separate test from the one on this page.
$\lim_{x\to c^{-}}f=\lim_{x\to c^{+}}f$Condition 2 written out. For a piecewise function this is the check you actually perform at the seam.
Removable discontinuityThe limit exists but condition 1 or 3 fails. Redefining $f(c)$ repairs it.
Non-removableCondition 2 fails — no choice of $f(c)$ can help, because there is nothing to match.
Worked examples
EX 1
Run the three-condition test
piecewiseLet $f(x)=\begin{cases}\dfrac{x^{2}-4}{x-2},& x\neq2\\ 5,& x=2\end{cases}$. Is $f$ continuous at $x=2$?
1
Condition 1
$f(2)=5$ is given, so the value exists.
2
Condition 2
For $x\neq2$, $\dfrac{(x-2)(x+2)}{x-2}=x+2$, so $\lim_{x\to2}f=4$. The limit exists.
3
Condition 3
$4\neq5$, so the limit and the value are not equal.
4
Conclude
$f$ is discontinuous at $x=2$. Because the limit exists, it is removable — redefining $f(2)=4$ would fix it.
5
How to write it
Name all three checks explicitly. “The limit is 4 but $f(2)=5$” is the sentence that earns the point.
EX 2
Solve for continuity
find $k$For which $k$ is $g(x)=\begin{cases}kx+1,& x\le 3\\ x^{2}-2,& x\gt 3\end{cases}$ continuous at $x=3$?
1
Left limit and value
From the top piece, $\lim_{x\to3^{-}}g=3k+1$, and $g(3)=3k+1$ as well.
2
Right limit
From the bottom piece, $\lim_{x\to3^{+}}g=9-2=7$.
3
Force condition 2
The one-sided limits must agree: $3k+1=7$.
5
Check condition 3
With $k=2$, $g(3)=7$ and the limit is $7$. All three conditions hold, so $g$ is continuous at $3$.
EX 3
Which condition failed?
diagnoseFor $h(x)=\dfrac{x+1}{x^{2}-1}$, classify the behaviour at $x=1$ and at $x=-1$.
1
Factor
$h(x)=\dfrac{x+1}{(x-1)(x+1)}=\dfrac{1}{x-1}$ for $x\neq\pm1$.
2
At $x=-1$
$h(-1)$ is undefined, so condition 1 fails. But $\lim_{x\to-1}h=\dfrac{1}{-2}=-\tfrac12$ exists — a removable discontinuity.
3
At $x=1$
$h(1)$ is undefined and $\dfrac{1}{x-1}$ blows up: the one-sided limits are $-\infty$ and $+\infty$. Condition 2 fails — non-removable (infinite).
4
The lesson
Both points look identical in the original formula (denominator zero). Only factoring reveals that one factor cancels and the other does not.
EX 4
Continuity does not imply differentiability
boundaryIs $f(x)=\lvert x\rvert$ continuous at $x=0$? Is it differentiable there?
1
Condition 1
$f(0)=0$ exists.
2
Condition 2
$\lim_{x\to0^{-}}\lvert x\rvert=0$ and $\lim_{x\to0^{+}}\lvert x\rvert=0$; they agree, so the limit is $0$.
3
Condition 3
$0=0$. All three hold, so $f$ is continuous at $0$.
4
Now the derivative
The slope is $-1$ on the left and $+1$ on the right, so $f'(0)$ does not exist — a corner.
5
State the relationship
Differentiable $\Rightarrow$ continuous, but not the converse. This example is the standard counterexample.
Boundary cases
Three conditions, not one
Writing only “$\lim f=f(c)$” without checking that each side exists is an incomplete justification, even when the conclusion is right.
Removable vs non-removable turns on condition 2
If the limit exists, the discontinuity is removable no matter which of conditions 1 or 3 failed. If the limit does not exist, nothing can repair it.
A piecewise seam is not automatically a discontinuity
You must check. Many piecewise functions are perfectly continuous at the join — Example 2 with $k=2$.
Continuous does not mean smooth
$\lvert x\rvert$ is continuous everywhere and differentiable nowhere at $0$. Do not use the words interchangeably.